Prove that inequality $frac2aba+b+sqrtfraca^2+b^22ge sqrtab+fraca+b2$

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Let $a;bge 0$. Prove that inequality $$frac2aba+b+sqrtfraca^2+b^22ge sqrtab+fraca+b2$$

---

My try: $LHS-RHS=frac2aba+b-fraca+b2+sqrtfraca^2+b^22-sqrtabge 0$

Or $-fracleft(a-bright)^22left(a+bright)+fracleft(a-bright)^22left(sqrtfraca^2+b^22+sqrtabright)ge 0$

Or $left(a-bright)^2left(frac12left(sqrtfraca^2+b^22+sqrtabright)-frac12left(a+bright)right)ge 0$

Then i can't prove $frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$

I squared the two sides of the inequality but the exponential is hard to solve. And I can see $HM+QMge AM+GM$ with $n=2$ so is that true for $n=i$?

inequality radicals fractions a.m.-g.m.-inequality cauchy-schwarz-inequality

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edited 20 hours ago

Michael Rozenberg

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asked 21 hours ago

Nguyễn Duy Linh

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up vote
0
down vote

favorite

1

Let $a;bge 0$. Prove that inequality $$frac2aba+b+sqrtfraca^2+b^22ge sqrtab+fraca+b2$$

---

My try: $LHS-RHS=frac2aba+b-fraca+b2+sqrtfraca^2+b^22-sqrtabge 0$

Or $-fracleft(a-bright)^22left(a+bright)+fracleft(a-bright)^22left(sqrtfraca^2+b^22+sqrtabright)ge 0$

Or $left(a-bright)^2left(frac12left(sqrtfraca^2+b^22+sqrtabright)-frac12left(a+bright)right)ge 0$

Then i can't prove $frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$

I squared the two sides of the inequality but the exponential is hard to solve. And I can see $HM+QMge AM+GM$ with $n=2$ so is that true for $n=i$?

inequality radicals fractions a.m.-g.m.-inequality cauchy-schwarz-inequality

share|cite|improve this question

edited 20 hours ago

Michael Rozenberg

86.9k1575178

asked 21 hours ago

Nguyễn Duy Linh

204

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

Let $a;bge 0$. Prove that inequality $$frac2aba+b+sqrtfraca^2+b^22ge sqrtab+fraca+b2$$

---

My try: $LHS-RHS=frac2aba+b-fraca+b2+sqrtfraca^2+b^22-sqrtabge 0$

Or $-fracleft(a-bright)^22left(a+bright)+fracleft(a-bright)^22left(sqrtfraca^2+b^22+sqrtabright)ge 0$

Or $left(a-bright)^2left(frac12left(sqrtfraca^2+b^22+sqrtabright)-frac12left(a+bright)right)ge 0$

Then i can't prove $frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$

I squared the two sides of the inequality but the exponential is hard to solve. And I can see $HM+QMge AM+GM$ with $n=2$ so is that true for $n=i$?

inequality radicals fractions a.m.-g.m.-inequality cauchy-schwarz-inequality

share|cite|improve this question

edited 20 hours ago

Michael Rozenberg

86.9k1575178

asked 21 hours ago

Nguyễn Duy Linh

204

Let $a;bge 0$. Prove that inequality $$frac2aba+b+sqrtfraca^2+b^22ge sqrtab+fraca+b2$$

---

My try: $LHS-RHS=frac2aba+b-fraca+b2+sqrtfraca^2+b^22-sqrtabge 0$

Or $-fracleft(a-bright)^22left(a+bright)+fracleft(a-bright)^22left(sqrtfraca^2+b^22+sqrtabright)ge 0$

Or $left(a-bright)^2left(frac12left(sqrtfraca^2+b^22+sqrtabright)-frac12left(a+bright)right)ge 0$

Then i can't prove $frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$

I squared the two sides of the inequality but the exponential is hard to solve. And I can see $HM+QMge AM+GM$ with $n=2$ so is that true for $n=i$?

inequality radicals fractions a.m.-g.m.-inequality cauchy-schwarz-inequality

share|cite|improve this question

edited 20 hours ago

Michael Rozenberg

86.9k1575178

asked 21 hours ago

Nguyễn Duy Linh

204

share|cite|improve this question

share|cite|improve this question

edited 20 hours ago

Michael Rozenberg

86.9k1575178

edited 20 hours ago

Michael Rozenberg

86.9k1575178

edited 20 hours ago

Michael Rozenberg

86.9k1575178

86.9k1575178

asked 21 hours ago

Nguyễn Duy Linh

204

asked 21 hours ago

Nguyễn Duy Linh

204

asked 21 hours ago

Nguyễn Duy Linh

204

204

add a comment | 

add a comment | 

6 Answers
6

active

oldest

votes

up vote
2
down vote



accepted

Note that $$beginalignsqrtfraca^2+b^22-sqrtab&=fracleft(fraca^2+b^22right)-absqrtfraca^2+b^22+sqrtab=frac2left(fraca-b2right)^2sqrtfraca^2+b^22+sqrtab
endalign$$
If we can show that
$$sqrtfraca^2+b^22+sqrtableq a+b,,$$
then
$$sqrtfraca^2+b^22-sqrtabgeq frac2left(fraca-b2right)^2a+b=fraca+b2-frac2aba+b,,$$
which is equivalent to what we need to prove.

Note that $$beginalign
sqrtfraca^2+b^22&=sqrtleft(fraca+b2right)^2+left(fraca-b2right)^2
\
&=sqrtleft(fraca+b2right)^2+left(fracsqrta-sqrtbsqrt2right)^2,left(fracsqrta+sqrtbsqrt2right)^2
\
&=sqrtleft(fraca+b2right)^2+left(fracsqrta-sqrtbsqrt2right)^2left(left(fracsqrta-sqrtbsqrt2right)^2+2sqrtabright)
\
&=sqrtleft(fraca+b2right)^2+2sqrtableft(fracsqrta-sqrtbsqrt2right)^2+left(fracsqrta-sqrtbsqrt2right)^4
\
&leqsqrtleft(fraca+b2right)^2+2left(fraca+b2right)left(fracsqrta-sqrtbsqrt2right)^2+left(fracsqrta-sqrtbsqrt2right)^4
\
&=fraca+b2+left(fracsqrta-sqrtbsqrt2right)^2,.endalign$$
Thus,
$$sqrtfraca^2+b^22+sqrtableq a+b,.$$

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edited 19 hours ago

answered 21 hours ago

Batominovski

22.2k22676

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  I think you can use Cauchy-Schwarz's inequality for prove $sqrtfraca^2+b^22+sqrtableq a+b,.$ it's easy than yours, btw thank you.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unless a Cauchy-Schwarz proof of that inequality is illustrated somewhere, I remain unconvinced. If such a proof doesn't exist, then it is certainly easier than my proof.
  – Batominovski
  21 hours ago
  
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  By C-S: $left(sqrtfraca^2+b^22+sqrtabright)^2le 2left(fraca^2+b^22+abright)=left(a+bright)^2$ Done :).
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good. You can write this as an answer, and I will upvote it. One can answer his own question.
  – Batominovski
  21 hours ago
  
  

  
  
  
  

add a comment | 

up vote
2
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This Problem is well known.
Note that $$fraca+b2-frac2aba+b=frac(a+b)^2-4ab2(a+b)$$
after squaring two times you will get

$$left(frac(a+b)^22-frac(a-b)^44(a+b)^2right)^2geq 4ableft(fraca^2+b^22right)$$
Doing this Algebra you will get

$$frac116frac(a-b)^8(a+b)^4geq 0$$ which is true.

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edited 21 hours ago

answered 21 hours ago

Dr. Sonnhard Graubner

66.6k32659

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  Thanks Dr.Son now i know your solution.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  

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up vote
2
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An alternative approach, let

$$
u = fraca + b2 quad textand quad v = ab
$$

so that the inequality becomes

$$
fracvu + sqrt2u^2 - v ge sqrtv + u.
$$

Since the original inequality is homogeneous, and clearly true for $a = b = 0$, we can without loss of generality set $u = 1$. Then letting $v = 1 + x$ reduces the inequality to

$$
x ge sqrt1 + x - sqrt1 - x quad (textfor,, x le 0)
$$

which is shown fairly easily. The condition comes from

$$
u = 1 implies b = 2 - a implies x = a(2 - a) - 1 = -(a-1)^2 le 0.
$$

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answered 21 hours ago

D. G.

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  Nice solutions are written here!
  – Dr. Sonnhard Graubner
  21 hours ago
  

  
  
  
  

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up vote
1
down vote

I think this is the simplest solution:

So you are stuck here:

$$frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$$

...which is equivalent to:

$$sqrtfraca^2+b^22+sqrtable a+btag1$$

Apply Jensen inequality:

$$fracf(x_1)+...+f(x_n)nle f(fracx_1+...+x_nn)$$

...for concave function $f(x)=sqrtx$ and $x_1=fraca^2+b^22$, $x_2=ab$:

$$fracsqrtfraca^2+b^22+sqrtab2le sqrtfracfraca^2+b^22+ab2tag2$$

Simplify (2) and you get (1) directly.

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answered 21 hours ago

Oldboy

2,3881216

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  That is a good solution too.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


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  Upvote won't hurt :)
  – Oldboy
  21 hours ago
  

  
  
  
  


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  Sorry but votes cast by those with less than 15 reputation are recorded .-. .
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


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  @NguyễnDuyLinh As I suggested, posy your own solution as an answer. I will upvote it. Then, you will get more than 15 reputation points.
  – Batominovski
  21 hours ago
  

  
  
  
  

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up vote
1
down vote

Thanks everyone i had the answer for my stuck.

We have : $$LHS-RHS=frac2aba+b-fraca+b2+sqrtfraca^2+b^22-sqrtabge 0$$

Or $$-fracleft(a-bright)^22left(a+bright)+fracleft(a-bright)^22left(sqrtfraca^2+b^22+sqrtabright)ge 0$$

Or $$left(a-bright)^2left(frac12left(sqrtfraca^2+b^22+sqrtabright)-frac12left(a+bright)right)ge 0$$

Then i need to prove $$frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$$

Or $sqrtfraca^2+b^22+sqrtableq a+b$. Which's true by C-S $$left(sqrtfraca^2+b^22+sqrtabright)^2le 2left(fraca^2+b^22+abright)=left(a+bright)^2$$

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answered 21 hours ago

Nguyễn Duy Linh

204

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  This is one of few cases that the OP answered his own question, and quite beautifully. An upvote is very well deserved.
  – Batominovski
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

Let $a^2+b^2=2t^2ab$, where $t>0$.

Thus, by AM-GM $tgeq1$ and we need to prove that
$$frac2absqrta^2+b^2+2ab+sqrtfraca^2+b^22geqsqrtab+fracsqrta^2+b^2+2ab2$$ or
$$sqrtfrac2t^2+1+tgeq1+sqrtfract^2+12$$ or
$$t-1geqsqrtfract^2+12-sqrtfrac2t^2+1$$ or
$$t-1geqfract^2-1sqrt2(t^2+1)$$ or
$$sqrt2(t^2+1)geq t+1,$$ which is true by C-S:
$$sqrt2(t^2+1)=sqrt(1^2+1^2)(t^2+1)geq t+1.$$
Done!

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answered 20 hours ago

Michael Rozenberg

86.9k1575178

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Note that $$beginalignsqrtfraca^2+b^22-sqrtab&=fracleft(fraca^2+b^22right)-absqrtfraca^2+b^22+sqrtab=frac2left(fraca-b2right)^2sqrtfraca^2+b^22+sqrtab
endalign$$
If we can show that
$$sqrtfraca^2+b^22+sqrtableq a+b,,$$
then
$$sqrtfraca^2+b^22-sqrtabgeq frac2left(fraca-b2right)^2a+b=fraca+b2-frac2aba+b,,$$
which is equivalent to what we need to prove.

Note that $$beginalign
sqrtfraca^2+b^22&=sqrtleft(fraca+b2right)^2+left(fraca-b2right)^2
\
&=sqrtleft(fraca+b2right)^2+left(fracsqrta-sqrtbsqrt2right)^2,left(fracsqrta+sqrtbsqrt2right)^2
\
&=sqrtleft(fraca+b2right)^2+left(fracsqrta-sqrtbsqrt2right)^2left(left(fracsqrta-sqrtbsqrt2right)^2+2sqrtabright)
\
&=sqrtleft(fraca+b2right)^2+2sqrtableft(fracsqrta-sqrtbsqrt2right)^2+left(fracsqrta-sqrtbsqrt2right)^4
\
&leqsqrtleft(fraca+b2right)^2+2left(fraca+b2right)left(fracsqrta-sqrtbsqrt2right)^2+left(fracsqrta-sqrtbsqrt2right)^4
\
&=fraca+b2+left(fracsqrta-sqrtbsqrt2right)^2,.endalign$$
Thus,
$$sqrtfraca^2+b^22+sqrtableq a+b,.$$

share|cite|improve this answer

edited 19 hours ago

answered 21 hours ago

Batominovski

22.2k22676

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you can use Cauchy-Schwarz's inequality for prove $sqrtfraca^2+b^22+sqrtableq a+b,.$ it's easy than yours, btw thank you.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unless a Cauchy-Schwarz proof of that inequality is illustrated somewhere, I remain unconvinced. If such a proof doesn't exist, then it is certainly easier than my proof.
  – Batominovski
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  By C-S: $left(sqrtfraca^2+b^22+sqrtabright)^2le 2left(fraca^2+b^22+abright)=left(a+bright)^2$ Done :).
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good. You can write this as an answer, and I will upvote it. One can answer his own question.
  – Batominovski
  21 hours ago
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Note that $$beginalignsqrtfraca^2+b^22-sqrtab&=fracleft(fraca^2+b^22right)-absqrtfraca^2+b^22+sqrtab=frac2left(fraca-b2right)^2sqrtfraca^2+b^22+sqrtab
endalign$$
If we can show that
$$sqrtfraca^2+b^22+sqrtableq a+b,,$$
then
$$sqrtfraca^2+b^22-sqrtabgeq frac2left(fraca-b2right)^2a+b=fraca+b2-frac2aba+b,,$$
which is equivalent to what we need to prove.

Note that $$beginalign
sqrtfraca^2+b^22&=sqrtleft(fraca+b2right)^2+left(fraca-b2right)^2
\
&=sqrtleft(fraca+b2right)^2+left(fracsqrta-sqrtbsqrt2right)^2,left(fracsqrta+sqrtbsqrt2right)^2
\
&=sqrtleft(fraca+b2right)^2+left(fracsqrta-sqrtbsqrt2right)^2left(left(fracsqrta-sqrtbsqrt2right)^2+2sqrtabright)
\
&=sqrtleft(fraca+b2right)^2+2sqrtableft(fracsqrta-sqrtbsqrt2right)^2+left(fracsqrta-sqrtbsqrt2right)^4
\
&leqsqrtleft(fraca+b2right)^2+2left(fraca+b2right)left(fracsqrta-sqrtbsqrt2right)^2+left(fracsqrta-sqrtbsqrt2right)^4
\
&=fraca+b2+left(fracsqrta-sqrtbsqrt2right)^2,.endalign$$
Thus,
$$sqrtfraca^2+b^22+sqrtableq a+b,.$$

share|cite|improve this answer

edited 19 hours ago

answered 21 hours ago

Batominovski

22.2k22676

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you can use Cauchy-Schwarz's inequality for prove $sqrtfraca^2+b^22+sqrtableq a+b,.$ it's easy than yours, btw thank you.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unless a Cauchy-Schwarz proof of that inequality is illustrated somewhere, I remain unconvinced. If such a proof doesn't exist, then it is certainly easier than my proof.
  – Batominovski
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  By C-S: $left(sqrtfraca^2+b^22+sqrtabright)^2le 2left(fraca^2+b^22+abright)=left(a+bright)^2$ Done :).
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good. You can write this as an answer, and I will upvote it. One can answer his own question.
  – Batominovski
  21 hours ago
  
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Note that $$beginalignsqrtfraca^2+b^22-sqrtab&=fracleft(fraca^2+b^22right)-absqrtfraca^2+b^22+sqrtab=frac2left(fraca-b2right)^2sqrtfraca^2+b^22+sqrtab
endalign$$
If we can show that
$$sqrtfraca^2+b^22+sqrtableq a+b,,$$
then
$$sqrtfraca^2+b^22-sqrtabgeq frac2left(fraca-b2right)^2a+b=fraca+b2-frac2aba+b,,$$
which is equivalent to what we need to prove.

Note that $$beginalign
sqrtfraca^2+b^22&=sqrtleft(fraca+b2right)^2+left(fraca-b2right)^2
\
&=sqrtleft(fraca+b2right)^2+left(fracsqrta-sqrtbsqrt2right)^2,left(fracsqrta+sqrtbsqrt2right)^2
\
&=sqrtleft(fraca+b2right)^2+left(fracsqrta-sqrtbsqrt2right)^2left(left(fracsqrta-sqrtbsqrt2right)^2+2sqrtabright)
\
&=sqrtleft(fraca+b2right)^2+2sqrtableft(fracsqrta-sqrtbsqrt2right)^2+left(fracsqrta-sqrtbsqrt2right)^4
\
&leqsqrtleft(fraca+b2right)^2+2left(fraca+b2right)left(fracsqrta-sqrtbsqrt2right)^2+left(fracsqrta-sqrtbsqrt2right)^4
\
&=fraca+b2+left(fracsqrta-sqrtbsqrt2right)^2,.endalign$$
Thus,
$$sqrtfraca^2+b^22+sqrtableq a+b,.$$

share|cite|improve this answer

edited 19 hours ago

answered 21 hours ago

Batominovski

22.2k22676

Note that $$beginalignsqrtfraca^2+b^22-sqrtab&=fracleft(fraca^2+b^22right)-absqrtfraca^2+b^22+sqrtab=frac2left(fraca-b2right)^2sqrtfraca^2+b^22+sqrtab
endalign$$
If we can show that
$$sqrtfraca^2+b^22+sqrtableq a+b,,$$
then
$$sqrtfraca^2+b^22-sqrtabgeq frac2left(fraca-b2right)^2a+b=fraca+b2-frac2aba+b,,$$
which is equivalent to what we need to prove.

Note that $$beginalign
sqrtfraca^2+b^22&=sqrtleft(fraca+b2right)^2+left(fraca-b2right)^2
\
&=sqrtleft(fraca+b2right)^2+left(fracsqrta-sqrtbsqrt2right)^2,left(fracsqrta+sqrtbsqrt2right)^2
\
&=sqrtleft(fraca+b2right)^2+left(fracsqrta-sqrtbsqrt2right)^2left(left(fracsqrta-sqrtbsqrt2right)^2+2sqrtabright)
\
&=sqrtleft(fraca+b2right)^2+2sqrtableft(fracsqrta-sqrtbsqrt2right)^2+left(fracsqrta-sqrtbsqrt2right)^4
\
&leqsqrtleft(fraca+b2right)^2+2left(fraca+b2right)left(fracsqrta-sqrtbsqrt2right)^2+left(fracsqrta-sqrtbsqrt2right)^4
\
&=fraca+b2+left(fracsqrta-sqrtbsqrt2right)^2,.endalign$$
Thus,
$$sqrtfraca^2+b^22+sqrtableq a+b,.$$

share|cite|improve this answer

edited 19 hours ago

answered 21 hours ago

Batominovski

22.2k22676

share|cite|improve this answer

share|cite|improve this answer

edited 19 hours ago

edited 19 hours ago

edited 19 hours ago

answered 21 hours ago

Batominovski

22.2k22676

answered 21 hours ago

Batominovski

22.2k22676

answered 21 hours ago

Batominovski

22.2k22676

22.2k22676

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you can use Cauchy-Schwarz's inequality for prove $sqrtfraca^2+b^22+sqrtableq a+b,.$ it's easy than yours, btw thank you.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unless a Cauchy-Schwarz proof of that inequality is illustrated somewhere, I remain unconvinced. If such a proof doesn't exist, then it is certainly easier than my proof.
  – Batominovski
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  By C-S: $left(sqrtfraca^2+b^22+sqrtabright)^2le 2left(fraca^2+b^22+abright)=left(a+bright)^2$ Done :).
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good. You can write this as an answer, and I will upvote it. One can answer his own question.
  – Batominovski
  21 hours ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I think you can use Cauchy-Schwarz's inequality for prove $sqrtfraca^2+b^22+sqrtableq a+b,.$ it's easy than yours, btw thank you.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unless a Cauchy-Schwarz proof of that inequality is illustrated somewhere, I remain unconvinced. If such a proof doesn't exist, then it is certainly easier than my proof.
  – Batominovski
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  By C-S: $left(sqrtfraca^2+b^22+sqrtabright)^2le 2left(fraca^2+b^22+abright)=left(a+bright)^2$ Done :).
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Good. You can write this as an answer, and I will upvote it. One can answer his own question.
  – Batominovski
  21 hours ago
  
  

  
  
  
  

I think you can use Cauchy-Schwarz's inequality for prove $sqrtfraca^2+b^22+sqrtableq a+b,.$ it's easy than yours, btw thank you.
– Nguyễn Duy Linh
21 hours ago

I think you can use Cauchy-Schwarz's inequality for prove $sqrtfraca^2+b^22+sqrtableq a+b,.$ it's easy than yours, btw thank you.
– Nguyễn Duy Linh
21 hours ago

Unless a Cauchy-Schwarz proof of that inequality is illustrated somewhere, I remain unconvinced. If such a proof doesn't exist, then it is certainly easier than my proof.
– Batominovski
21 hours ago

Unless a Cauchy-Schwarz proof of that inequality is illustrated somewhere, I remain unconvinced. If such a proof doesn't exist, then it is certainly easier than my proof.
– Batominovski
21 hours ago

1

1

By C-S: $left(sqrtfraca^2+b^22+sqrtabright)^2le 2left(fraca^2+b^22+abright)=left(a+bright)^2$ Done :).
– Nguyễn Duy Linh
21 hours ago

By C-S: $left(sqrtfraca^2+b^22+sqrtabright)^2le 2left(fraca^2+b^22+abright)=left(a+bright)^2$ Done :).
– Nguyễn Duy Linh
21 hours ago

Good. You can write this as an answer, and I will upvote it. One can answer his own question.
– Batominovski
21 hours ago

Good. You can write this as an answer, and I will upvote it. One can answer his own question.
– Batominovski
21 hours ago

add a comment | 

up vote
2
down vote

This Problem is well known.
Note that $$fraca+b2-frac2aba+b=frac(a+b)^2-4ab2(a+b)$$
after squaring two times you will get

$$left(frac(a+b)^22-frac(a-b)^44(a+b)^2right)^2geq 4ableft(fraca^2+b^22right)$$
Doing this Algebra you will get

$$frac116frac(a-b)^8(a+b)^4geq 0$$ which is true.

share|cite|improve this answer

edited 21 hours ago

answered 21 hours ago

Dr. Sonnhard Graubner

66.6k32659

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Dr.Son now i know your solution.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

This Problem is well known.
Note that $$fraca+b2-frac2aba+b=frac(a+b)^2-4ab2(a+b)$$
after squaring two times you will get

$$left(frac(a+b)^22-frac(a-b)^44(a+b)^2right)^2geq 4ableft(fraca^2+b^22right)$$
Doing this Algebra you will get

$$frac116frac(a-b)^8(a+b)^4geq 0$$ which is true.

share|cite|improve this answer

edited 21 hours ago

answered 21 hours ago

Dr. Sonnhard Graubner

66.6k32659

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Dr.Son now i know your solution.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

This Problem is well known.
Note that $$fraca+b2-frac2aba+b=frac(a+b)^2-4ab2(a+b)$$
after squaring two times you will get

$$left(frac(a+b)^22-frac(a-b)^44(a+b)^2right)^2geq 4ableft(fraca^2+b^22right)$$
Doing this Algebra you will get

$$frac116frac(a-b)^8(a+b)^4geq 0$$ which is true.

share|cite|improve this answer

edited 21 hours ago

answered 21 hours ago

Dr. Sonnhard Graubner

66.6k32659

This Problem is well known.
Note that $$fraca+b2-frac2aba+b=frac(a+b)^2-4ab2(a+b)$$
after squaring two times you will get

$$left(frac(a+b)^22-frac(a-b)^44(a+b)^2right)^2geq 4ableft(fraca^2+b^22right)$$
Doing this Algebra you will get

$$frac116frac(a-b)^8(a+b)^4geq 0$$ which is true.

share|cite|improve this answer

edited 21 hours ago

answered 21 hours ago

Dr. Sonnhard Graubner

66.6k32659

share|cite|improve this answer

share|cite|improve this answer

edited 21 hours ago

edited 21 hours ago

edited 21 hours ago

answered 21 hours ago

Dr. Sonnhard Graubner

66.6k32659

answered 21 hours ago

Dr. Sonnhard Graubner

66.6k32659

answered 21 hours ago

Dr. Sonnhard Graubner

66.6k32659

66.6k32659

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Dr.Son now i know your solution.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks Dr.Son now i know your solution.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  

Thanks Dr.Son now i know your solution.
– Nguyễn Duy Linh
21 hours ago

Thanks Dr.Son now i know your solution.
– Nguyễn Duy Linh
21 hours ago

add a comment | 

up vote
2
down vote

An alternative approach, let

$$
u = fraca + b2 quad textand quad v = ab
$$

so that the inequality becomes

$$
fracvu + sqrt2u^2 - v ge sqrtv + u.
$$

Since the original inequality is homogeneous, and clearly true for $a = b = 0$, we can without loss of generality set $u = 1$. Then letting $v = 1 + x$ reduces the inequality to

$$
x ge sqrt1 + x - sqrt1 - x quad (textfor,, x le 0)
$$

which is shown fairly easily. The condition comes from

$$
u = 1 implies b = 2 - a implies x = a(2 - a) - 1 = -(a-1)^2 le 0.
$$

share|cite|improve this answer

answered 21 hours ago

D. G.

464

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice solutions are written here!
  – Dr. Sonnhard Graubner
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

An alternative approach, let

$$
u = fraca + b2 quad textand quad v = ab
$$

so that the inequality becomes

$$
fracvu + sqrt2u^2 - v ge sqrtv + u.
$$

Since the original inequality is homogeneous, and clearly true for $a = b = 0$, we can without loss of generality set $u = 1$. Then letting $v = 1 + x$ reduces the inequality to

$$
x ge sqrt1 + x - sqrt1 - x quad (textfor,, x le 0)
$$

which is shown fairly easily. The condition comes from

$$
u = 1 implies b = 2 - a implies x = a(2 - a) - 1 = -(a-1)^2 le 0.
$$

share|cite|improve this answer

answered 21 hours ago

D. G.

464

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice solutions are written here!
  – Dr. Sonnhard Graubner
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

An alternative approach, let

$$
u = fraca + b2 quad textand quad v = ab
$$

so that the inequality becomes

$$
fracvu + sqrt2u^2 - v ge sqrtv + u.
$$

Since the original inequality is homogeneous, and clearly true for $a = b = 0$, we can without loss of generality set $u = 1$. Then letting $v = 1 + x$ reduces the inequality to

$$
x ge sqrt1 + x - sqrt1 - x quad (textfor,, x le 0)
$$

which is shown fairly easily. The condition comes from

$$
u = 1 implies b = 2 - a implies x = a(2 - a) - 1 = -(a-1)^2 le 0.
$$

share|cite|improve this answer

answered 21 hours ago

D. G.

464

An alternative approach, let

$$
u = fraca + b2 quad textand quad v = ab
$$

so that the inequality becomes

$$
fracvu + sqrt2u^2 - v ge sqrtv + u.
$$

Since the original inequality is homogeneous, and clearly true for $a = b = 0$, we can without loss of generality set $u = 1$. Then letting $v = 1 + x$ reduces the inequality to

$$
x ge sqrt1 + x - sqrt1 - x quad (textfor,, x le 0)
$$

which is shown fairly easily. The condition comes from

$$
u = 1 implies b = 2 - a implies x = a(2 - a) - 1 = -(a-1)^2 le 0.
$$

share|cite|improve this answer

answered 21 hours ago

D. G.

464

share|cite|improve this answer

share|cite|improve this answer

answered 21 hours ago

D. G.

464

answered 21 hours ago

D. G.

464

answered 21 hours ago

D. G.

464

464

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice solutions are written here!
  – Dr. Sonnhard Graubner
  21 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nice solutions are written here!
  – Dr. Sonnhard Graubner
  21 hours ago
  

  
  
  
  

Nice solutions are written here!
– Dr. Sonnhard Graubner
21 hours ago

Nice solutions are written here!
– Dr. Sonnhard Graubner
21 hours ago

add a comment | 

up vote
1
down vote

I think this is the simplest solution:

So you are stuck here:

$$frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$$

...which is equivalent to:

$$sqrtfraca^2+b^22+sqrtable a+btag1$$

Apply Jensen inequality:

$$fracf(x_1)+...+f(x_n)nle f(fracx_1+...+x_nn)$$

...for concave function $f(x)=sqrtx$ and $x_1=fraca^2+b^22$, $x_2=ab$:

$$fracsqrtfraca^2+b^22+sqrtab2le sqrtfracfraca^2+b^22+ab2tag2$$

Simplify (2) and you get (1) directly.

share|cite|improve this answer

answered 21 hours ago

Oldboy

2,3881216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is a good solution too.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Upvote won't hurt :)
  – Oldboy
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry but votes cast by those with less than 15 reputation are recorded .-. .
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @NguyễnDuyLinh As I suggested, posy your own solution as an answer. I will upvote it. Then, you will get more than 15 reputation points.
  – Batominovski
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

I think this is the simplest solution:

So you are stuck here:

$$frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$$

...which is equivalent to:

$$sqrtfraca^2+b^22+sqrtable a+btag1$$

Apply Jensen inequality:

$$fracf(x_1)+...+f(x_n)nle f(fracx_1+...+x_nn)$$

...for concave function $f(x)=sqrtx$ and $x_1=fraca^2+b^22$, $x_2=ab$:

$$fracsqrtfraca^2+b^22+sqrtab2le sqrtfracfraca^2+b^22+ab2tag2$$

Simplify (2) and you get (1) directly.

share|cite|improve this answer

answered 21 hours ago

Oldboy

2,3881216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is a good solution too.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Upvote won't hurt :)
  – Oldboy
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry but votes cast by those with less than 15 reputation are recorded .-. .
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @NguyễnDuyLinh As I suggested, posy your own solution as an answer. I will upvote it. Then, you will get more than 15 reputation points.
  – Batominovski
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

I think this is the simplest solution:

So you are stuck here:

$$frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$$

...which is equivalent to:

$$sqrtfraca^2+b^22+sqrtable a+btag1$$

Apply Jensen inequality:

$$fracf(x_1)+...+f(x_n)nle f(fracx_1+...+x_nn)$$

...for concave function $f(x)=sqrtx$ and $x_1=fraca^2+b^22$, $x_2=ab$:

$$fracsqrtfraca^2+b^22+sqrtab2le sqrtfracfraca^2+b^22+ab2tag2$$

Simplify (2) and you get (1) directly.

share|cite|improve this answer

answered 21 hours ago

Oldboy

2,3881216

I think this is the simplest solution:

So you are stuck here:

$$frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$$

...which is equivalent to:

$$sqrtfraca^2+b^22+sqrtable a+btag1$$

Apply Jensen inequality:

$$fracf(x_1)+...+f(x_n)nle f(fracx_1+...+x_nn)$$

...for concave function $f(x)=sqrtx$ and $x_1=fraca^2+b^22$, $x_2=ab$:

$$fracsqrtfraca^2+b^22+sqrtab2le sqrtfracfraca^2+b^22+ab2tag2$$

Simplify (2) and you get (1) directly.

share|cite|improve this answer

answered 21 hours ago

Oldboy

2,3881216

share|cite|improve this answer

share|cite|improve this answer

answered 21 hours ago

Oldboy

2,3881216

answered 21 hours ago

Oldboy

2,3881216

answered 21 hours ago

Oldboy

2,3881216

2,3881216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is a good solution too.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Upvote won't hurt :)
  – Oldboy
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry but votes cast by those with less than 15 reputation are recorded .-. .
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @NguyễnDuyLinh As I suggested, posy your own solution as an answer. I will upvote it. Then, you will get more than 15 reputation points.
  – Batominovski
  21 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is a good solution too.
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Upvote won't hurt :)
  – Oldboy
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sorry but votes cast by those with less than 15 reputation are recorded .-. .
  – Nguyễn Duy Linh
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @NguyễnDuyLinh As I suggested, posy your own solution as an answer. I will upvote it. Then, you will get more than 15 reputation points.
  – Batominovski
  21 hours ago
  

  
  
  
  

That is a good solution too.
– Nguyễn Duy Linh
21 hours ago

That is a good solution too.
– Nguyễn Duy Linh
21 hours ago

Upvote won't hurt :)
– Oldboy
21 hours ago

Upvote won't hurt :)
– Oldboy
21 hours ago

Sorry but votes cast by those with less than 15 reputation are recorded .-. .
– Nguyễn Duy Linh
21 hours ago

Sorry but votes cast by those with less than 15 reputation are recorded .-. .
– Nguyễn Duy Linh
21 hours ago

@NguyễnDuyLinh As I suggested, posy your own solution as an answer. I will upvote it. Then, you will get more than 15 reputation points.
– Batominovski
21 hours ago

@NguyễnDuyLinh As I suggested, posy your own solution as an answer. I will upvote it. Then, you will get more than 15 reputation points.
– Batominovski
21 hours ago

add a comment | 

up vote
1
down vote

Thanks everyone i had the answer for my stuck.

We have : $$LHS-RHS=frac2aba+b-fraca+b2+sqrtfraca^2+b^22-sqrtabge 0$$

Or $$-fracleft(a-bright)^22left(a+bright)+fracleft(a-bright)^22left(sqrtfraca^2+b^22+sqrtabright)ge 0$$

Or $$left(a-bright)^2left(frac12left(sqrtfraca^2+b^22+sqrtabright)-frac12left(a+bright)right)ge 0$$

Then i need to prove $$frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$$

Or $sqrtfraca^2+b^22+sqrtableq a+b$. Which's true by C-S $$left(sqrtfraca^2+b^22+sqrtabright)^2le 2left(fraca^2+b^22+abright)=left(a+bright)^2$$

share|cite|improve this answer

answered 21 hours ago

Nguyễn Duy Linh

204

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is one of few cases that the OP answered his own question, and quite beautifully. An upvote is very well deserved.
  – Batominovski
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Thanks everyone i had the answer for my stuck.

We have : $$LHS-RHS=frac2aba+b-fraca+b2+sqrtfraca^2+b^22-sqrtabge 0$$

Or $$-fracleft(a-bright)^22left(a+bright)+fracleft(a-bright)^22left(sqrtfraca^2+b^22+sqrtabright)ge 0$$

Or $$left(a-bright)^2left(frac12left(sqrtfraca^2+b^22+sqrtabright)-frac12left(a+bright)right)ge 0$$

Then i need to prove $$frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$$

Or $sqrtfraca^2+b^22+sqrtableq a+b$. Which's true by C-S $$left(sqrtfraca^2+b^22+sqrtabright)^2le 2left(fraca^2+b^22+abright)=left(a+bright)^2$$

share|cite|improve this answer

answered 21 hours ago

Nguyễn Duy Linh

204

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is one of few cases that the OP answered his own question, and quite beautifully. An upvote is very well deserved.
  – Batominovski
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Thanks everyone i had the answer for my stuck.

We have : $$LHS-RHS=frac2aba+b-fraca+b2+sqrtfraca^2+b^22-sqrtabge 0$$

Or $$-fracleft(a-bright)^22left(a+bright)+fracleft(a-bright)^22left(sqrtfraca^2+b^22+sqrtabright)ge 0$$

Or $$left(a-bright)^2left(frac12left(sqrtfraca^2+b^22+sqrtabright)-frac12left(a+bright)right)ge 0$$

Then i need to prove $$frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$$

Or $sqrtfraca^2+b^22+sqrtableq a+b$. Which's true by C-S $$left(sqrtfraca^2+b^22+sqrtabright)^2le 2left(fraca^2+b^22+abright)=left(a+bright)^2$$

share|cite|improve this answer

answered 21 hours ago

Nguyễn Duy Linh

204

Thanks everyone i had the answer for my stuck.

We have : $$LHS-RHS=frac2aba+b-fraca+b2+sqrtfraca^2+b^22-sqrtabge 0$$

Or $$-fracleft(a-bright)^22left(a+bright)+fracleft(a-bright)^22left(sqrtfraca^2+b^22+sqrtabright)ge 0$$

Or $$left(a-bright)^2left(frac12left(sqrtfraca^2+b^22+sqrtabright)-frac12left(a+bright)right)ge 0$$

Then i need to prove $$frac12left(sqrtfraca^2+b^22+sqrtabright)ge frac12left(a+bright)$$

Or $sqrtfraca^2+b^22+sqrtableq a+b$. Which's true by C-S $$left(sqrtfraca^2+b^22+sqrtabright)^2le 2left(fraca^2+b^22+abright)=left(a+bright)^2$$

share|cite|improve this answer

answered 21 hours ago

Nguyễn Duy Linh

204

share|cite|improve this answer

share|cite|improve this answer

answered 21 hours ago

Nguyễn Duy Linh

204

answered 21 hours ago

Nguyễn Duy Linh

204

answered 21 hours ago

Nguyễn Duy Linh

204

204

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is one of few cases that the OP answered his own question, and quite beautifully. An upvote is very well deserved.
  – Batominovski
  21 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  This is one of few cases that the OP answered his own question, and quite beautifully. An upvote is very well deserved.
  – Batominovski
  21 hours ago
  

  
  
  
  

1

1

This is one of few cases that the OP answered his own question, and quite beautifully. An upvote is very well deserved.
– Batominovski
21 hours ago

This is one of few cases that the OP answered his own question, and quite beautifully. An upvote is very well deserved.
– Batominovski
21 hours ago

add a comment | 

up vote
0
down vote

Let $a^2+b^2=2t^2ab$, where $t>0$.

Thus, by AM-GM $tgeq1$ and we need to prove that
$$frac2absqrta^2+b^2+2ab+sqrtfraca^2+b^22geqsqrtab+fracsqrta^2+b^2+2ab2$$ or
$$sqrtfrac2t^2+1+tgeq1+sqrtfract^2+12$$ or
$$t-1geqsqrtfract^2+12-sqrtfrac2t^2+1$$ or
$$t-1geqfract^2-1sqrt2(t^2+1)$$ or
$$sqrt2(t^2+1)geq t+1,$$ which is true by C-S:
$$sqrt2(t^2+1)=sqrt(1^2+1^2)(t^2+1)geq t+1.$$
Done!

share|cite|improve this answer

answered 20 hours ago

Michael Rozenberg

86.9k1575178

add a comment | 

up vote
0
down vote

Let $a^2+b^2=2t^2ab$, where $t>0$.

Thus, by AM-GM $tgeq1$ and we need to prove that
$$frac2absqrta^2+b^2+2ab+sqrtfraca^2+b^22geqsqrtab+fracsqrta^2+b^2+2ab2$$ or
$$sqrtfrac2t^2+1+tgeq1+sqrtfract^2+12$$ or
$$t-1geqsqrtfract^2+12-sqrtfrac2t^2+1$$ or
$$t-1geqfract^2-1sqrt2(t^2+1)$$ or
$$sqrt2(t^2+1)geq t+1,$$ which is true by C-S:
$$sqrt2(t^2+1)=sqrt(1^2+1^2)(t^2+1)geq t+1.$$
Done!

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answered 20 hours ago

Michael Rozenberg

86.9k1575178

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Let $a^2+b^2=2t^2ab$, where $t>0$.

Thus, by AM-GM $tgeq1$ and we need to prove that
$$frac2absqrta^2+b^2+2ab+sqrtfraca^2+b^22geqsqrtab+fracsqrta^2+b^2+2ab2$$ or
$$sqrtfrac2t^2+1+tgeq1+sqrtfract^2+12$$ or
$$t-1geqsqrtfract^2+12-sqrtfrac2t^2+1$$ or
$$t-1geqfract^2-1sqrt2(t^2+1)$$ or
$$sqrt2(t^2+1)geq t+1,$$ which is true by C-S:
$$sqrt2(t^2+1)=sqrt(1^2+1^2)(t^2+1)geq t+1.$$
Done!

share|cite|improve this answer

answered 20 hours ago

Michael Rozenberg

86.9k1575178

Let $a^2+b^2=2t^2ab$, where $t>0$.

Thus, by AM-GM $tgeq1$ and we need to prove that
$$frac2absqrta^2+b^2+2ab+sqrtfraca^2+b^22geqsqrtab+fracsqrta^2+b^2+2ab2$$ or
$$sqrtfrac2t^2+1+tgeq1+sqrtfract^2+12$$ or
$$t-1geqsqrtfract^2+12-sqrtfrac2t^2+1$$ or
$$t-1geqfract^2-1sqrt2(t^2+1)$$ or
$$sqrt2(t^2+1)geq t+1,$$ which is true by C-S:
$$sqrt2(t^2+1)=sqrt(1^2+1^2)(t^2+1)geq t+1.$$
Done!

share|cite|improve this answer

answered 20 hours ago

Michael Rozenberg

86.9k1575178

share|cite|improve this answer

share|cite|improve this answer

answered 20 hours ago

Michael Rozenberg

86.9k1575178

answered 20 hours ago

Michael Rozenberg

86.9k1575178

answered 20 hours ago

Michael Rozenberg

86.9k1575178

86.9k1575178

add a comment | 

add a comment | 

 

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