Prove $a^4+b^4+1ge a+b$

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up vote
9
down vote

favorite

3

Prove for every real $a,b$:
$$a^4+b^4+1ge a+b$$

What I've tried:

1.I checked how AM-GM may help but doesn't look like it's useful here.

2. I've tried:

$$(a^2+b^2)^2 -2(ab)^2+1 ge a+b$$

But unfortunately, I can't find any way to continue this..
I'm sure that this isn't too hard, it's just that "I'm not seeing it", I would appreciate if clues are given first so I can answer this myself.

*This exercise is from the TAU entry exams.

algebra-precalculus inequality

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asked Jul 20 at 20:28

Maxim

616314

add a comment | 

up vote
9
down vote

favorite

3

Prove for every real $a,b$:
$$a^4+b^4+1ge a+b$$

What I've tried:

1.I checked how AM-GM may help but doesn't look like it's useful here.

2. I've tried:

$$(a^2+b^2)^2 -2(ab)^2+1 ge a+b$$

But unfortunately, I can't find any way to continue this..
I'm sure that this isn't too hard, it's just that "I'm not seeing it", I would appreciate if clues are given first so I can answer this myself.

*This exercise is from the TAU entry exams.

algebra-precalculus inequality

share|cite|improve this question

asked Jul 20 at 20:28

Maxim

616314

add a comment | 

up vote
9
down vote

favorite

3

up vote
9
down vote

favorite

3

3

Prove for every real $a,b$:
$$a^4+b^4+1ge a+b$$

What I've tried:

1.I checked how AM-GM may help but doesn't look like it's useful here.

2. I've tried:

$$(a^2+b^2)^2 -2(ab)^2+1 ge a+b$$

But unfortunately, I can't find any way to continue this..
I'm sure that this isn't too hard, it's just that "I'm not seeing it", I would appreciate if clues are given first so I can answer this myself.

*This exercise is from the TAU entry exams.

algebra-precalculus inequality

share|cite|improve this question

asked Jul 20 at 20:28

Maxim

616314

Prove for every real $a,b$:
$$a^4+b^4+1ge a+b$$

What I've tried:

1.I checked how AM-GM may help but doesn't look like it's useful here.

2. I've tried:

$$(a^2+b^2)^2 -2(ab)^2+1 ge a+b$$

But unfortunately, I can't find any way to continue this..
I'm sure that this isn't too hard, it's just that "I'm not seeing it", I would appreciate if clues are given first so I can answer this myself.

*This exercise is from the TAU entry exams.

algebra-precalculus inequality

share|cite|improve this question

asked Jul 20 at 20:28

Maxim

616314

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 20:28

Maxim

616314

asked Jul 20 at 20:28

Maxim

616314

asked Jul 20 at 20:28

Maxim

616314

616314

add a comment | 

add a comment | 

5 Answers
5

active

oldest

votes

up vote
14
down vote



accepted

Proof

Since $$left(a^4-a^2+frac14right)+left(b^4-b^2+frac14right)=left(a^2-frac12right)^2+left(b^2-frac12right)^2 geq0,$$
hence $$a^4+b^4+1 geq a^2+b^2+frac12.tag1$$

Since $$left(a^2-a+frac14right)+left(b^2-b+frac14right)=left(a-frac12right)^2+left(b-frac12right)^2 geq 0,$$
hence, $$a^2+b^2+frac12 geq a+b.tag2$$

Combining $(1)$ and $(2)$, $$a^4+b^4+1 geq a+b.$$

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edited Jul 20 at 22:33

answered Jul 20 at 22:07

mengdie1982

2,912216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please elaborate a bit more? I'm having trouble to understand the logic between each step.
  – Maxim
  Jul 21 at 12:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  this is a purely elementary proof，which just brings in an intermediate variable. Can you agree that $x^2-2xy+y^2=(x-y)^2 geq 0$?
  – mengdie1982
  Jul 21 at 14:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes of course I can see that, It's just that it's hard for me to see how the first row of the proof leads to the second - that's the only thing.
  – Maxim
  Jul 22 at 8:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Maxim Do you know the transposition of the inequality？ For example， $x-y geq 0$ may imply $x geq y$.
  – mengdie1982
  Jul 22 at 14:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ok, sure, that makes sense but the part that was strange for me is why the second step has a positive one half and not a negative, after looking at it one more time, I perfectly understand everything, indeed a very nice solution!
  – Maxim
  Jul 22 at 16:27
  

  
  
  
  

add a comment | 

up vote
15
down vote

Note that, by the AM-GM Inequality, $$x^4+frac12=x^4+3left(frac16right)geq 4sqrt[4]x^4left(frac16right)^3=sqrt[4]frac256216,|x|geq |x|geq x$$ for all $xinmathbbR$. The equality does not hold, though.

---

A sharper inequality is $a^4+b^4+frac3sqrt[3]2^5geq a+b$. This is because
$$x^4+frac3sqrt[3]2^8 =x^4+3left(frac1sqrt[3]2^8right)geq 4sqrt[4]x^4left(frac1sqrt[3]2^8right)^3=|x|geq x,.$$
The inequality above is an equality if and only if $x=frac1sqrt[3]2^2$.

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edited Jul 21 at 8:02

answered Jul 20 at 20:33

Batominovski

23.2k22777

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you solve it so quickly? i.e. what ideas came to your mind when you read it?
  – Yousef Kaddoura
  Jul 20 at 20:35
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  The first idea was to try to separate the variables. So, when I saw the problem statement, I immediately thought that the inequalities $a^4+frac12geq a$ and $b^4+frac12geq b$ should hold.
  – Batominovski
  Jul 20 at 20:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks; I will keep this tactic in mind.
  – Yousef Kaddoura
  Jul 20 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $$x^4 + 3(frac16) geq 4(x^4(frac16)^3)^1/4$$ How did you do that using $$sumlimits_i=1^n a_ilambda_i geq Pi a_i^lambda_i$$ I could see that $lambda_2 = 3$ and $lambda_1 = 1$ But where did the 4th root come from ? Thanks
  – Ahmad Bazzi
  Jul 20 at 21:11
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $$x^4,frac16,frac16,frac16$$ are four numbers.
  – Dr. Sonnhard Graubner
  Jul 20 at 21:18
  

  
  
  
  

 | 
show 1 more comment

up vote
6
down vote

We have

$$a^4+b^4+1ge a+biff a^4-a+b^4-b+1ge 0$$

and it is easy to show that

$$f(x)=x^4-ximplies f'(x)=4x^3-1implies x_min=frac14^frac13quad f(x)ge f(x_min)approx-0.4725$$

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answered Jul 20 at 20:40

gimusi

65.4k73584

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could work with $f(x)=x^4-x+frac 12$ to preserve the symmetry and since $f(x)>0$ then $f(a)+f(b)$ too. Why keep the $1$ apart ?
  – zwim
  Jul 20 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zwim Yes of course w ecan also do in that way, it is almost equivalent.
  – gimusi
  Jul 20 at 21:41
  

  
  
  
  

add a comment | 

up vote
3
down vote

Clues only in this answer, since you said you want to figure out the answer yourself.

Have you tried proving it for positive numbers greater than $2$? The square of such a number is obviously greater than $2$, and much more so the biquadrate. Prove it for those numbers and the answer is obvious for negative numbers less than $-2$.

Where it gets really tricky is in the unit interval. Let's say $$a = b = frac12.$$ Then $$left(frac12right)^4 = frac116,$$ so $$a^4 + b^4 = frac18,$$ which is actually less than either $a$ or $b$ alone. But then you get to add $1$ to that...

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answered Jul 20 at 20:41

The Short One

7601621

add a comment | 

up vote
2
down vote

Let's have $4c^3=1$ and $f(a)=a^4-a+frac 12$

$beginalignrequirecancelf(c+u)-f(c) &=cancelc^4+cancel4c^3u+6c^2u^2+4cu^3+u^4-cancelc-cancelu+cancelfrac 12-cancelc^4+cancelc-cancelfrac 12\
&=u^2underbrace(u^2+4cu+6c^2)_Delta=-8c^2<0ge 0endalign$

So $f(c)$ is a minimum and $f(a)ge f(c)approx0.0275>0$

The conclusion arises from $f(a)+f(b)>0$.

share|cite|improve this answer

answered Jul 21 at 1:58

zwim

11k627

add a comment | 

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
14
down vote



accepted

Proof

Since $$left(a^4-a^2+frac14right)+left(b^4-b^2+frac14right)=left(a^2-frac12right)^2+left(b^2-frac12right)^2 geq0,$$
hence $$a^4+b^4+1 geq a^2+b^2+frac12.tag1$$

Since $$left(a^2-a+frac14right)+left(b^2-b+frac14right)=left(a-frac12right)^2+left(b-frac12right)^2 geq 0,$$
hence, $$a^2+b^2+frac12 geq a+b.tag2$$

Combining $(1)$ and $(2)$, $$a^4+b^4+1 geq a+b.$$

share|cite|improve this answer

edited Jul 20 at 22:33

answered Jul 20 at 22:07

mengdie1982

2,912216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please elaborate a bit more? I'm having trouble to understand the logic between each step.
  – Maxim
  Jul 21 at 12:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  this is a purely elementary proof，which just brings in an intermediate variable. Can you agree that $x^2-2xy+y^2=(x-y)^2 geq 0$?
  – mengdie1982
  Jul 21 at 14:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes of course I can see that, It's just that it's hard for me to see how the first row of the proof leads to the second - that's the only thing.
  – Maxim
  Jul 22 at 8:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Maxim Do you know the transposition of the inequality？ For example， $x-y geq 0$ may imply $x geq y$.
  – mengdie1982
  Jul 22 at 14:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ok, sure, that makes sense but the part that was strange for me is why the second step has a positive one half and not a negative, after looking at it one more time, I perfectly understand everything, indeed a very nice solution!
  – Maxim
  Jul 22 at 16:27
  

  
  
  
  

add a comment | 

up vote
14
down vote



accepted

Proof

Since $$left(a^4-a^2+frac14right)+left(b^4-b^2+frac14right)=left(a^2-frac12right)^2+left(b^2-frac12right)^2 geq0,$$
hence $$a^4+b^4+1 geq a^2+b^2+frac12.tag1$$

Since $$left(a^2-a+frac14right)+left(b^2-b+frac14right)=left(a-frac12right)^2+left(b-frac12right)^2 geq 0,$$
hence, $$a^2+b^2+frac12 geq a+b.tag2$$

Combining $(1)$ and $(2)$, $$a^4+b^4+1 geq a+b.$$

share|cite|improve this answer

edited Jul 20 at 22:33

answered Jul 20 at 22:07

mengdie1982

2,912216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please elaborate a bit more? I'm having trouble to understand the logic between each step.
  – Maxim
  Jul 21 at 12:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  this is a purely elementary proof，which just brings in an intermediate variable. Can you agree that $x^2-2xy+y^2=(x-y)^2 geq 0$?
  – mengdie1982
  Jul 21 at 14:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes of course I can see that, It's just that it's hard for me to see how the first row of the proof leads to the second - that's the only thing.
  – Maxim
  Jul 22 at 8:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Maxim Do you know the transposition of the inequality？ For example， $x-y geq 0$ may imply $x geq y$.
  – mengdie1982
  Jul 22 at 14:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ok, sure, that makes sense but the part that was strange for me is why the second step has a positive one half and not a negative, after looking at it one more time, I perfectly understand everything, indeed a very nice solution!
  – Maxim
  Jul 22 at 16:27
  

  
  
  
  

add a comment | 

up vote
14
down vote



accepted

up vote
14
down vote



accepted

Proof

Since $$left(a^4-a^2+frac14right)+left(b^4-b^2+frac14right)=left(a^2-frac12right)^2+left(b^2-frac12right)^2 geq0,$$
hence $$a^4+b^4+1 geq a^2+b^2+frac12.tag1$$

Since $$left(a^2-a+frac14right)+left(b^2-b+frac14right)=left(a-frac12right)^2+left(b-frac12right)^2 geq 0,$$
hence, $$a^2+b^2+frac12 geq a+b.tag2$$

Combining $(1)$ and $(2)$, $$a^4+b^4+1 geq a+b.$$

share|cite|improve this answer

edited Jul 20 at 22:33

answered Jul 20 at 22:07

mengdie1982

2,912216

Proof

Since $$left(a^4-a^2+frac14right)+left(b^4-b^2+frac14right)=left(a^2-frac12right)^2+left(b^2-frac12right)^2 geq0,$$
hence $$a^4+b^4+1 geq a^2+b^2+frac12.tag1$$

Since $$left(a^2-a+frac14right)+left(b^2-b+frac14right)=left(a-frac12right)^2+left(b-frac12right)^2 geq 0,$$
hence, $$a^2+b^2+frac12 geq a+b.tag2$$

Combining $(1)$ and $(2)$, $$a^4+b^4+1 geq a+b.$$

share|cite|improve this answer

edited Jul 20 at 22:33

answered Jul 20 at 22:07

mengdie1982

2,912216

share|cite|improve this answer

share|cite|improve this answer

edited Jul 20 at 22:33

edited Jul 20 at 22:33

edited Jul 20 at 22:33

answered Jul 20 at 22:07

mengdie1982

2,912216

answered Jul 20 at 22:07

mengdie1982

2,912216

answered Jul 20 at 22:07

mengdie1982

2,912216

2,912216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please elaborate a bit more? I'm having trouble to understand the logic between each step.
  – Maxim
  Jul 21 at 12:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  this is a purely elementary proof，which just brings in an intermediate variable. Can you agree that $x^2-2xy+y^2=(x-y)^2 geq 0$?
  – mengdie1982
  Jul 21 at 14:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes of course I can see that, It's just that it's hard for me to see how the first row of the proof leads to the second - that's the only thing.
  – Maxim
  Jul 22 at 8:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Maxim Do you know the transposition of the inequality？ For example， $x-y geq 0$ may imply $x geq y$.
  – mengdie1982
  Jul 22 at 14:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ok, sure, that makes sense but the part that was strange for me is why the second step has a positive one half and not a negative, after looking at it one more time, I perfectly understand everything, indeed a very nice solution!
  – Maxim
  Jul 22 at 16:27
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please elaborate a bit more? I'm having trouble to understand the logic between each step.
  – Maxim
  Jul 21 at 12:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  this is a purely elementary proof，which just brings in an intermediate variable. Can you agree that $x^2-2xy+y^2=(x-y)^2 geq 0$?
  – mengdie1982
  Jul 21 at 14:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes of course I can see that, It's just that it's hard for me to see how the first row of the proof leads to the second - that's the only thing.
  – Maxim
  Jul 22 at 8:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Maxim Do you know the transposition of the inequality？ For example， $x-y geq 0$ may imply $x geq y$.
  – mengdie1982
  Jul 22 at 14:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ok, sure, that makes sense but the part that was strange for me is why the second step has a positive one half and not a negative, after looking at it one more time, I perfectly understand everything, indeed a very nice solution!
  – Maxim
  Jul 22 at 16:27
  

  
  
  
  

Can you please elaborate a bit more? I'm having trouble to understand the logic between each step.
– Maxim
Jul 21 at 12:26

Can you please elaborate a bit more? I'm having trouble to understand the logic between each step.
– Maxim
Jul 21 at 12:26

1

1

this is a purely elementary proof，which just brings in an intermediate variable. Can you agree that $x^2-2xy+y^2=(x-y)^2 geq 0$?
– mengdie1982
Jul 21 at 14:30

this is a purely elementary proof，which just brings in an intermediate variable. Can you agree that $x^2-2xy+y^2=(x-y)^2 geq 0$?
– mengdie1982
Jul 21 at 14:30

Yes of course I can see that, It's just that it's hard for me to see how the first row of the proof leads to the second - that's the only thing.
– Maxim
Jul 22 at 8:06

Yes of course I can see that, It's just that it's hard for me to see how the first row of the proof leads to the second - that's the only thing.
– Maxim
Jul 22 at 8:06

@Maxim Do you know the transposition of the inequality？ For example， $x-y geq 0$ may imply $x geq y$.
– mengdie1982
Jul 22 at 14:37

@Maxim Do you know the transposition of the inequality？ For example， $x-y geq 0$ may imply $x geq y$.
– mengdie1982
Jul 22 at 14:37

ok, sure, that makes sense but the part that was strange for me is why the second step has a positive one half and not a negative, after looking at it one more time, I perfectly understand everything, indeed a very nice solution!
– Maxim
Jul 22 at 16:27

ok, sure, that makes sense but the part that was strange for me is why the second step has a positive one half and not a negative, after looking at it one more time, I perfectly understand everything, indeed a very nice solution!
– Maxim
Jul 22 at 16:27

add a comment | 

up vote
15
down vote

Note that, by the AM-GM Inequality, $$x^4+frac12=x^4+3left(frac16right)geq 4sqrt[4]x^4left(frac16right)^3=sqrt[4]frac256216,|x|geq |x|geq x$$ for all $xinmathbbR$. The equality does not hold, though.

---

A sharper inequality is $a^4+b^4+frac3sqrt[3]2^5geq a+b$. This is because
$$x^4+frac3sqrt[3]2^8 =x^4+3left(frac1sqrt[3]2^8right)geq 4sqrt[4]x^4left(frac1sqrt[3]2^8right)^3=|x|geq x,.$$
The inequality above is an equality if and only if $x=frac1sqrt[3]2^2$.

share|cite|improve this answer

edited Jul 21 at 8:02

answered Jul 20 at 20:33

Batominovski

23.2k22777

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you solve it so quickly? i.e. what ideas came to your mind when you read it?
  – Yousef Kaddoura
  Jul 20 at 20:35
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  The first idea was to try to separate the variables. So, when I saw the problem statement, I immediately thought that the inequalities $a^4+frac12geq a$ and $b^4+frac12geq b$ should hold.
  – Batominovski
  Jul 20 at 20:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks; I will keep this tactic in mind.
  – Yousef Kaddoura
  Jul 20 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $$x^4 + 3(frac16) geq 4(x^4(frac16)^3)^1/4$$ How did you do that using $$sumlimits_i=1^n a_ilambda_i geq Pi a_i^lambda_i$$ I could see that $lambda_2 = 3$ and $lambda_1 = 1$ But where did the 4th root come from ? Thanks
  – Ahmad Bazzi
  Jul 20 at 21:11
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $$x^4,frac16,frac16,frac16$$ are four numbers.
  – Dr. Sonnhard Graubner
  Jul 20 at 21:18
  

  
  
  
  

 | 
show 1 more comment

up vote
15
down vote

Note that, by the AM-GM Inequality, $$x^4+frac12=x^4+3left(frac16right)geq 4sqrt[4]x^4left(frac16right)^3=sqrt[4]frac256216,|x|geq |x|geq x$$ for all $xinmathbbR$. The equality does not hold, though.

---

A sharper inequality is $a^4+b^4+frac3sqrt[3]2^5geq a+b$. This is because
$$x^4+frac3sqrt[3]2^8 =x^4+3left(frac1sqrt[3]2^8right)geq 4sqrt[4]x^4left(frac1sqrt[3]2^8right)^3=|x|geq x,.$$
The inequality above is an equality if and only if $x=frac1sqrt[3]2^2$.

share|cite|improve this answer

edited Jul 21 at 8:02

answered Jul 20 at 20:33

Batominovski

23.2k22777

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you solve it so quickly? i.e. what ideas came to your mind when you read it?
  – Yousef Kaddoura
  Jul 20 at 20:35
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  The first idea was to try to separate the variables. So, when I saw the problem statement, I immediately thought that the inequalities $a^4+frac12geq a$ and $b^4+frac12geq b$ should hold.
  – Batominovski
  Jul 20 at 20:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks; I will keep this tactic in mind.
  – Yousef Kaddoura
  Jul 20 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $$x^4 + 3(frac16) geq 4(x^4(frac16)^3)^1/4$$ How did you do that using $$sumlimits_i=1^n a_ilambda_i geq Pi a_i^lambda_i$$ I could see that $lambda_2 = 3$ and $lambda_1 = 1$ But where did the 4th root come from ? Thanks
  – Ahmad Bazzi
  Jul 20 at 21:11
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $$x^4,frac16,frac16,frac16$$ are four numbers.
  – Dr. Sonnhard Graubner
  Jul 20 at 21:18
  

  
  
  
  

 | 
show 1 more comment

up vote
15
down vote

up vote
15
down vote

Note that, by the AM-GM Inequality, $$x^4+frac12=x^4+3left(frac16right)geq 4sqrt[4]x^4left(frac16right)^3=sqrt[4]frac256216,|x|geq |x|geq x$$ for all $xinmathbbR$. The equality does not hold, though.

---

A sharper inequality is $a^4+b^4+frac3sqrt[3]2^5geq a+b$. This is because
$$x^4+frac3sqrt[3]2^8 =x^4+3left(frac1sqrt[3]2^8right)geq 4sqrt[4]x^4left(frac1sqrt[3]2^8right)^3=|x|geq x,.$$
The inequality above is an equality if and only if $x=frac1sqrt[3]2^2$.

share|cite|improve this answer

edited Jul 21 at 8:02

answered Jul 20 at 20:33

Batominovski

23.2k22777

Note that, by the AM-GM Inequality, $$x^4+frac12=x^4+3left(frac16right)geq 4sqrt[4]x^4left(frac16right)^3=sqrt[4]frac256216,|x|geq |x|geq x$$ for all $xinmathbbR$. The equality does not hold, though.

---

A sharper inequality is $a^4+b^4+frac3sqrt[3]2^5geq a+b$. This is because
$$x^4+frac3sqrt[3]2^8 =x^4+3left(frac1sqrt[3]2^8right)geq 4sqrt[4]x^4left(frac1sqrt[3]2^8right)^3=|x|geq x,.$$
The inequality above is an equality if and only if $x=frac1sqrt[3]2^2$.

share|cite|improve this answer

edited Jul 21 at 8:02

answered Jul 20 at 20:33

Batominovski

23.2k22777

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 8:02

edited Jul 21 at 8:02

edited Jul 21 at 8:02

answered Jul 20 at 20:33

Batominovski

23.2k22777

answered Jul 20 at 20:33

Batominovski

23.2k22777

answered Jul 20 at 20:33

Batominovski

23.2k22777

23.2k22777

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you solve it so quickly? i.e. what ideas came to your mind when you read it?
  – Yousef Kaddoura
  Jul 20 at 20:35
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  The first idea was to try to separate the variables. So, when I saw the problem statement, I immediately thought that the inequalities $a^4+frac12geq a$ and $b^4+frac12geq b$ should hold.
  – Batominovski
  Jul 20 at 20:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks; I will keep this tactic in mind.
  – Yousef Kaddoura
  Jul 20 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $$x^4 + 3(frac16) geq 4(x^4(frac16)^3)^1/4$$ How did you do that using $$sumlimits_i=1^n a_ilambda_i geq Pi a_i^lambda_i$$ I could see that $lambda_2 = 3$ and $lambda_1 = 1$ But where did the 4th root come from ? Thanks
  – Ahmad Bazzi
  Jul 20 at 21:11
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $$x^4,frac16,frac16,frac16$$ are four numbers.
  – Dr. Sonnhard Graubner
  Jul 20 at 21:18
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How did you solve it so quickly? i.e. what ideas came to your mind when you read it?
  – Yousef Kaddoura
  Jul 20 at 20:35
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  The first idea was to try to separate the variables. So, when I saw the problem statement, I immediately thought that the inequalities $a^4+frac12geq a$ and $b^4+frac12geq b$ should hold.
  – Batominovski
  Jul 20 at 20:37
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks; I will keep this tactic in mind.
  – Yousef Kaddoura
  Jul 20 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $$x^4 + 3(frac16) geq 4(x^4(frac16)^3)^1/4$$ How did you do that using $$sumlimits_i=1^n a_ilambda_i geq Pi a_i^lambda_i$$ I could see that $lambda_2 = 3$ and $lambda_1 = 1$ But where did the 4th root come from ? Thanks
  – Ahmad Bazzi
  Jul 20 at 21:11
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $$x^4,frac16,frac16,frac16$$ are four numbers.
  – Dr. Sonnhard Graubner
  Jul 20 at 21:18
  

  
  
  
  

How did you solve it so quickly? i.e. what ideas came to your mind when you read it?
– Yousef Kaddoura
Jul 20 at 20:35

How did you solve it so quickly? i.e. what ideas came to your mind when you read it?
– Yousef Kaddoura
Jul 20 at 20:35

5

5

The first idea was to try to separate the variables. So, when I saw the problem statement, I immediately thought that the inequalities $a^4+frac12geq a$ and $b^4+frac12geq b$ should hold.
– Batominovski
Jul 20 at 20:37

The first idea was to try to separate the variables. So, when I saw the problem statement, I immediately thought that the inequalities $a^4+frac12geq a$ and $b^4+frac12geq b$ should hold.
– Batominovski
Jul 20 at 20:37

Thanks; I will keep this tactic in mind.
– Yousef Kaddoura
Jul 20 at 20:38

Thanks; I will keep this tactic in mind.
– Yousef Kaddoura
Jul 20 at 20:38

$$x^4 + 3(frac16) geq 4(x^4(frac16)^3)^1/4$$ How did you do that using $$sumlimits_i=1^n a_ilambda_i geq Pi a_i^lambda_i$$ I could see that $lambda_2 = 3$ and $lambda_1 = 1$ But where did the 4th root come from ? Thanks
– Ahmad Bazzi
Jul 20 at 21:11

$$x^4 + 3(frac16) geq 4(x^4(frac16)^3)^1/4$$ How did you do that using $$sumlimits_i=1^n a_ilambda_i geq Pi a_i^lambda_i$$ I could see that $lambda_2 = 3$ and $lambda_1 = 1$ But where did the 4th root come from ? Thanks
– Ahmad Bazzi
Jul 20 at 21:11

1

1

$$x^4,frac16,frac16,frac16$$ are four numbers.
– Dr. Sonnhard Graubner
Jul 20 at 21:18

$$x^4,frac16,frac16,frac16$$ are four numbers.
– Dr. Sonnhard Graubner
Jul 20 at 21:18

 | 
show 1 more comment

up vote
6
down vote

We have

$$a^4+b^4+1ge a+biff a^4-a+b^4-b+1ge 0$$

and it is easy to show that

$$f(x)=x^4-ximplies f'(x)=4x^3-1implies x_min=frac14^frac13quad f(x)ge f(x_min)approx-0.4725$$

share|cite|improve this answer

answered Jul 20 at 20:40

gimusi

65.4k73584

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could work with $f(x)=x^4-x+frac 12$ to preserve the symmetry and since $f(x)>0$ then $f(a)+f(b)$ too. Why keep the $1$ apart ?
  – zwim
  Jul 20 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zwim Yes of course w ecan also do in that way, it is almost equivalent.
  – gimusi
  Jul 20 at 21:41
  

  
  
  
  

add a comment | 

up vote
6
down vote

We have

$$a^4+b^4+1ge a+biff a^4-a+b^4-b+1ge 0$$

and it is easy to show that

$$f(x)=x^4-ximplies f'(x)=4x^3-1implies x_min=frac14^frac13quad f(x)ge f(x_min)approx-0.4725$$

share|cite|improve this answer

answered Jul 20 at 20:40

gimusi

65.4k73584

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could work with $f(x)=x^4-x+frac 12$ to preserve the symmetry and since $f(x)>0$ then $f(a)+f(b)$ too. Why keep the $1$ apart ?
  – zwim
  Jul 20 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zwim Yes of course w ecan also do in that way, it is almost equivalent.
  – gimusi
  Jul 20 at 21:41
  

  
  
  
  

add a comment | 

up vote
6
down vote

up vote
6
down vote

We have

$$a^4+b^4+1ge a+biff a^4-a+b^4-b+1ge 0$$

and it is easy to show that

$$f(x)=x^4-ximplies f'(x)=4x^3-1implies x_min=frac14^frac13quad f(x)ge f(x_min)approx-0.4725$$

share|cite|improve this answer

answered Jul 20 at 20:40

gimusi

65.4k73584

We have

$$a^4+b^4+1ge a+biff a^4-a+b^4-b+1ge 0$$

and it is easy to show that

$$f(x)=x^4-ximplies f'(x)=4x^3-1implies x_min=frac14^frac13quad f(x)ge f(x_min)approx-0.4725$$

share|cite|improve this answer

answered Jul 20 at 20:40

gimusi

65.4k73584

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 20:40

gimusi

65.4k73584

answered Jul 20 at 20:40

gimusi

65.4k73584

answered Jul 20 at 20:40

gimusi

65.4k73584

65.4k73584

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could work with $f(x)=x^4-x+frac 12$ to preserve the symmetry and since $f(x)>0$ then $f(a)+f(b)$ too. Why keep the $1$ apart ?
  – zwim
  Jul 20 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zwim Yes of course w ecan also do in that way, it is almost equivalent.
  – gimusi
  Jul 20 at 21:41
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could work with $f(x)=x^4-x+frac 12$ to preserve the symmetry and since $f(x)>0$ then $f(a)+f(b)$ too. Why keep the $1$ apart ?
  – zwim
  Jul 20 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @zwim Yes of course w ecan also do in that way, it is almost equivalent.
  – gimusi
  Jul 20 at 21:41
  

  
  
  
  

You could work with $f(x)=x^4-x+frac 12$ to preserve the symmetry and since $f(x)>0$ then $f(a)+f(b)$ too. Why keep the $1$ apart ?
– zwim
Jul 20 at 21:37

You could work with $f(x)=x^4-x+frac 12$ to preserve the symmetry and since $f(x)>0$ then $f(a)+f(b)$ too. Why keep the $1$ apart ?
– zwim
Jul 20 at 21:37

@zwim Yes of course w ecan also do in that way, it is almost equivalent.
– gimusi
Jul 20 at 21:41

@zwim Yes of course w ecan also do in that way, it is almost equivalent.
– gimusi
Jul 20 at 21:41

add a comment | 

up vote
3
down vote

Clues only in this answer, since you said you want to figure out the answer yourself.

Have you tried proving it for positive numbers greater than $2$? The square of such a number is obviously greater than $2$, and much more so the biquadrate. Prove it for those numbers and the answer is obvious for negative numbers less than $-2$.

Where it gets really tricky is in the unit interval. Let's say $$a = b = frac12.$$ Then $$left(frac12right)^4 = frac116,$$ so $$a^4 + b^4 = frac18,$$ which is actually less than either $a$ or $b$ alone. But then you get to add $1$ to that...

share|cite|improve this answer

answered Jul 20 at 20:41

The Short One

7601621

add a comment | 

up vote
3
down vote

Clues only in this answer, since you said you want to figure out the answer yourself.

Have you tried proving it for positive numbers greater than $2$? The square of such a number is obviously greater than $2$, and much more so the biquadrate. Prove it for those numbers and the answer is obvious for negative numbers less than $-2$.

Where it gets really tricky is in the unit interval. Let's say $$a = b = frac12.$$ Then $$left(frac12right)^4 = frac116,$$ so $$a^4 + b^4 = frac18,$$ which is actually less than either $a$ or $b$ alone. But then you get to add $1$ to that...

share|cite|improve this answer

answered Jul 20 at 20:41

The Short One

7601621

add a comment | 

up vote
3
down vote

up vote
3
down vote

Clues only in this answer, since you said you want to figure out the answer yourself.

Have you tried proving it for positive numbers greater than $2$? The square of such a number is obviously greater than $2$, and much more so the biquadrate. Prove it for those numbers and the answer is obvious for negative numbers less than $-2$.

Where it gets really tricky is in the unit interval. Let's say $$a = b = frac12.$$ Then $$left(frac12right)^4 = frac116,$$ so $$a^4 + b^4 = frac18,$$ which is actually less than either $a$ or $b$ alone. But then you get to add $1$ to that...

share|cite|improve this answer

answered Jul 20 at 20:41

The Short One

7601621

Clues only in this answer, since you said you want to figure out the answer yourself.

Have you tried proving it for positive numbers greater than $2$? The square of such a number is obviously greater than $2$, and much more so the biquadrate. Prove it for those numbers and the answer is obvious for negative numbers less than $-2$.

Where it gets really tricky is in the unit interval. Let's say $$a = b = frac12.$$ Then $$left(frac12right)^4 = frac116,$$ so $$a^4 + b^4 = frac18,$$ which is actually less than either $a$ or $b$ alone. But then you get to add $1$ to that...

share|cite|improve this answer

answered Jul 20 at 20:41

The Short One

7601621

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 20:41

The Short One

7601621

answered Jul 20 at 20:41

The Short One

7601621

answered Jul 20 at 20:41

The Short One

7601621

7601621

add a comment | 

add a comment | 

up vote
2
down vote

Let's have $4c^3=1$ and $f(a)=a^4-a+frac 12$

$beginalignrequirecancelf(c+u)-f(c) &=cancelc^4+cancel4c^3u+6c^2u^2+4cu^3+u^4-cancelc-cancelu+cancelfrac 12-cancelc^4+cancelc-cancelfrac 12\
&=u^2underbrace(u^2+4cu+6c^2)_Delta=-8c^2<0ge 0endalign$

So $f(c)$ is a minimum and $f(a)ge f(c)approx0.0275>0$

The conclusion arises from $f(a)+f(b)>0$.

share|cite|improve this answer

answered Jul 21 at 1:58

zwim

11k627

add a comment | 

up vote
2
down vote

Let's have $4c^3=1$ and $f(a)=a^4-a+frac 12$

$beginalignrequirecancelf(c+u)-f(c) &=cancelc^4+cancel4c^3u+6c^2u^2+4cu^3+u^4-cancelc-cancelu+cancelfrac 12-cancelc^4+cancelc-cancelfrac 12\
&=u^2underbrace(u^2+4cu+6c^2)_Delta=-8c^2<0ge 0endalign$

So $f(c)$ is a minimum and $f(a)ge f(c)approx0.0275>0$

The conclusion arises from $f(a)+f(b)>0$.

share|cite|improve this answer

answered Jul 21 at 1:58

zwim

11k627

add a comment | 

up vote
2
down vote

up vote
2
down vote

Let's have $4c^3=1$ and $f(a)=a^4-a+frac 12$

$beginalignrequirecancelf(c+u)-f(c) &=cancelc^4+cancel4c^3u+6c^2u^2+4cu^3+u^4-cancelc-cancelu+cancelfrac 12-cancelc^4+cancelc-cancelfrac 12\
&=u^2underbrace(u^2+4cu+6c^2)_Delta=-8c^2<0ge 0endalign$

So $f(c)$ is a minimum and $f(a)ge f(c)approx0.0275>0$

The conclusion arises from $f(a)+f(b)>0$.

share|cite|improve this answer

answered Jul 21 at 1:58

zwim

11k627

Let's have $4c^3=1$ and $f(a)=a^4-a+frac 12$

$beginalignrequirecancelf(c+u)-f(c) &=cancelc^4+cancel4c^3u+6c^2u^2+4cu^3+u^4-cancelc-cancelu+cancelfrac 12-cancelc^4+cancelc-cancelfrac 12\
&=u^2underbrace(u^2+4cu+6c^2)_Delta=-8c^2<0ge 0endalign$

So $f(c)$ is a minimum and $f(a)ge f(c)approx0.0275>0$

The conclusion arises from $f(a)+f(b)>0$.

share|cite|improve this answer

answered Jul 21 at 1:58

zwim

11k627

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 1:58

zwim

11k627

answered Jul 21 at 1:58

zwim

11k627

answered Jul 21 at 1:58

zwim

11k627

11k627

add a comment | 

add a comment | 

 

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