Proving $frac1sin(A/2)+frac1sin(B/2)+frac1sin(C/2)ge 6$, where $A$, $B$, $C$ are angles of a triangle [duplicate]

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  Proving $cscfracalpha2+cscfracbeta2+cscfracgamma2 ge 6$, where $alpha$, $beta$, $gamma$ are the angles of a triangle [closed]
  
  2 answers
  
  

If $A$, $B$, and $C$ are the angles of a triangle, then $$frac1sin left(fracA2right)+frac1sinleft(fracB2right)+frac1sinleft(fracC2right)ge 6$$

I have used multiple trigonometric identities, but the situation becomes complicated. I also thought about the Sine Law. To be honest, I don’t think these techniques are suitable. Any suggestions?

geometry trigonometry inequality euclidean-geometry geometric-inequalities

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edited Jul 26 at 8:56

Michael Rozenberg

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asked Jul 25 at 20:11

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Jul 25 at 21:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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  Martin R ...may i ask a question : How did you find that my question is duplicate i have searched the site many times????
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  With Approach0: approach0.xyz/search/… – See also math.meta.stackexchange.com/questions/24978/….
  – Martin R
  Jul 25 at 20:45
  
  

  
  
  
  

add a comment | 

up vote
3
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This question already has an answer here:

• 
  Proving $cscfracalpha2+cscfracbeta2+cscfracgamma2 ge 6$, where $alpha$, $beta$, $gamma$ are the angles of a triangle [closed]
  
  2 answers
  
  

If $A$, $B$, and $C$ are the angles of a triangle, then $$frac1sin left(fracA2right)+frac1sinleft(fracB2right)+frac1sinleft(fracC2right)ge 6$$

I have used multiple trigonometric identities, but the situation becomes complicated. I also thought about the Sine Law. To be honest, I don’t think these techniques are suitable. Any suggestions?

geometry trigonometry inequality euclidean-geometry geometric-inequalities

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edited Jul 26 at 8:56

Michael Rozenberg

87.8k1578180

asked Jul 25 at 20:11

Johnny chad

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marked as duplicate by Martin R, Arnaud Mortier, Robert Z, B. Mehta, Blue geometry
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Jul 25 at 21:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Martin R ...may i ask a question : How did you find that my question is duplicate i have searched the site many times????
  – Johnny chad
  Jul 25 at 20:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  With Approach0: approach0.xyz/search/… – See also math.meta.stackexchange.com/questions/24978/….
  – Martin R
  Jul 25 at 20:45
  
  

  
  
  
  

add a comment | 

up vote
3
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favorite

up vote
3
down vote

favorite

This question already has an answer here:

• 
  Proving $cscfracalpha2+cscfracbeta2+cscfracgamma2 ge 6$, where $alpha$, $beta$, $gamma$ are the angles of a triangle [closed]
  
  2 answers
  
  

If $A$, $B$, and $C$ are the angles of a triangle, then $$frac1sin left(fracA2right)+frac1sinleft(fracB2right)+frac1sinleft(fracC2right)ge 6$$

I have used multiple trigonometric identities, but the situation becomes complicated. I also thought about the Sine Law. To be honest, I don’t think these techniques are suitable. Any suggestions?

geometry trigonometry inequality euclidean-geometry geometric-inequalities

share|cite|improve this question

edited Jul 26 at 8:56

Michael Rozenberg

87.8k1578180

asked Jul 25 at 20:11

Johnny chad

1106

This question already has an answer here:

• 
  Proving $cscfracalpha2+cscfracbeta2+cscfracgamma2 ge 6$, where $alpha$, $beta$, $gamma$ are the angles of a triangle [closed]
  
  2 answers
  
  

If $A$, $B$, and $C$ are the angles of a triangle, then $$frac1sin left(fracA2right)+frac1sinleft(fracB2right)+frac1sinleft(fracC2right)ge 6$$

I have used multiple trigonometric identities, but the situation becomes complicated. I also thought about the Sine Law. To be honest, I don’t think these techniques are suitable. Any suggestions?

This question already has an answer here:

• 
  Proving $cscfracalpha2+cscfracbeta2+cscfracgamma2 ge 6$, where $alpha$, $beta$, $gamma$ are the angles of a triangle [closed]
  
  2 answers
  
  

geometry trigonometry inequality euclidean-geometry geometric-inequalities

share|cite|improve this question

edited Jul 26 at 8:56

Michael Rozenberg

87.8k1578180

asked Jul 25 at 20:11

Johnny chad

1106

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 8:56

Michael Rozenberg

87.8k1578180

edited Jul 26 at 8:56

Michael Rozenberg

87.8k1578180

edited Jul 26 at 8:56

Michael Rozenberg

87.8k1578180

87.8k1578180

asked Jul 25 at 20:11

Johnny chad

1106

asked Jul 25 at 20:11

Johnny chad

1106

asked Jul 25 at 20:11

Johnny chad

1106

1106

marked as duplicate by Martin R, Arnaud Mortier, Robert Z, B. Mehta, Blue geometry
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Jul 25 at 21:30

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Jul 25 at 21:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Martin R ...may i ask a question : How did you find that my question is duplicate i have searched the site many times????
  – Johnny chad
  Jul 25 at 20:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  With Approach0: approach0.xyz/search/… – See also math.meta.stackexchange.com/questions/24978/….
  – Martin R
  Jul 25 at 20:45
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Martin R ...may i ask a question : How did you find that my question is duplicate i have searched the site many times????
  – Johnny chad
  Jul 25 at 20:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  With Approach0: approach0.xyz/search/… – See also math.meta.stackexchange.com/questions/24978/….
  – Martin R
  Jul 25 at 20:45
  
  

  
  
  
  

Martin R ...may i ask a question : How did you find that my question is duplicate i have searched the site many times????
– Johnny chad
Jul 25 at 20:39

Martin R ...may i ask a question : How did you find that my question is duplicate i have searched the site many times????
– Johnny chad
Jul 25 at 20:39

With Approach0: approach0.xyz/search/… – See also math.meta.stackexchange.com/questions/24978/….
– Martin R
Jul 25 at 20:45

With Approach0: approach0.xyz/search/… – See also math.meta.stackexchange.com/questions/24978/….
– Martin R
Jul 25 at 20:45

add a comment | 

6 Answers
6

active

oldest

votes

up vote
3
down vote



accepted

HINT:

Consider
$$fleft( x right)=1/sin left( fracx2 right);,quad 0<x<pi $$

It easy to verify that $f''left( x right)>0$ and we have a convex function. Now Using Jensen’s inequality:
$$frac13fleft( fracA2 right)+frac13fleft( fracB2 right)+frac13fleft( fracC2 right)ge fleft( frac13fracA2+frac13fracB2+frac13fracC2 right)$$

share|cite|improve this answer

answered Jul 25 at 20:23

Vincent Law

1,457212

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks man this is simple after some show you the solution
  – Johnny chad
  Jul 25 at 20:37
  

  
  
  
  

add a comment | 

up vote
3
down vote

A Pure Geometric Proof

Let $I$ be the incenter of $triangle ABC$, $r$ be the radius of the incircle. Then $$dfracrsindfracA2+dfracrsindfracB2+dfracrsindfracC2=IA+IB+IC.tag1$$

Now, let's apply Erdos-Mordell Inequality，which states that:

from a point $P$ inside a given $triangle ABC$， the perpendiculars $PX,
PY, PZ$ are drawn to its sides. Then

$$PA+PB+PC≥2(PX+PY+PZ).$$

Thus, we instantly have

$$IA+IB+ICgeq 2(r+r+r)=6r.tag2$$
Combining $(1)$ and $(2)$, we readily obtain what we want to prove.

share|cite|improve this answer

edited Jul 26 at 7:39

answered Jul 25 at 21:20

mengdie1982

2,835216

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up vote
2
down vote

We have that

$$f(x)=frac1sin x$$

is convex then by Jensen's inequality

$$fracfrac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)3ge frac1sinleft(fracA+B+C6right)$$

and then

$$frac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)ge frac3sinleft(fracA+B+C6right)=6$$

share|cite|improve this answer

answered Jul 25 at 20:22

gimusi

65k73583

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up vote
0
down vote

Also, by AM-GM in the standard notation we obtain:
$$sum_cycfrac1sinfracalpha2geqfrac3sqrt[3]prodlimits_cycsinfracalpha2=frac3sqrt[3]fracr4Rgeqfrac3sqrt[3]frac14cdot2=6.$$

share|cite|improve this answer

answered Jul 25 at 20:40

Michael Rozenberg

87.8k1578180

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down vote?
  – Michael Rozenberg
  Jul 28 at 3:32
  

  
  
  
  

add a comment | 

up vote
0
down vote

We use $$sinleft(fracA2right)=sqrtfrac(s-b)(s-c)bc$$ etc then we get by AM-GM:
$$frac13left(sqrtfracbc(s-b)(s-c)+sqrtfracac(s-a)(s-c)+sqrtfracab(s-a)(s-b)right)geq sqrt[3]fracabc(s-a)(s-b)(s-c)geq 2$$
if $$fracabc(s-a)(s-b)(s-c)geq 8$$

and this is $$abcgeq (-a+b+c)(a-b+c)(a+b-c)$$
using the Substitution

$$a=y+z,b=x+z,c=x+y$$

we get

$$(x+y)(x+z)(z+x)geq 8xyz$$ and this is AM-GM!
Note that $$s=fraca+b+c2$$

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answered Jul 25 at 20:45

Dr. Sonnhard Graubner

66.7k32659

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up vote
0
down vote

Applying Lagrange Multipliers shows that $$ sin(A/2) + sin (B/2) + sin (B/2)$$ is maximized when the triangle is equilateral in which case we have $$ sin(A/2) + sin (B/2) + sin (B/2)=3/2$$

The Arithmetic Harmonic inequality implies $$frac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)ge frac9sin(A/2) +sin(B/2)+sin(C/2)ge frac 9(3/2)=6$$

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answered Jul 25 at 20:47

Mohammad Riazi-Kermani

27.4k41852

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

HINT:

Consider
$$fleft( x right)=1/sin left( fracx2 right);,quad 0<x<pi $$

It easy to verify that $f''left( x right)>0$ and we have a convex function. Now Using Jensen’s inequality:
$$frac13fleft( fracA2 right)+frac13fleft( fracB2 right)+frac13fleft( fracC2 right)ge fleft( frac13fracA2+frac13fracB2+frac13fracC2 right)$$

share|cite|improve this answer

answered Jul 25 at 20:23

Vincent Law

1,457212

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks man this is simple after some show you the solution
  – Johnny chad
  Jul 25 at 20:37
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

HINT:

Consider
$$fleft( x right)=1/sin left( fracx2 right);,quad 0<x<pi $$

It easy to verify that $f''left( x right)>0$ and we have a convex function. Now Using Jensen’s inequality:
$$frac13fleft( fracA2 right)+frac13fleft( fracB2 right)+frac13fleft( fracC2 right)ge fleft( frac13fracA2+frac13fracB2+frac13fracC2 right)$$

share|cite|improve this answer

answered Jul 25 at 20:23

Vincent Law

1,457212

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks man this is simple after some show you the solution
  – Johnny chad
  Jul 25 at 20:37
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

HINT:

Consider
$$fleft( x right)=1/sin left( fracx2 right);,quad 0<x<pi $$

It easy to verify that $f''left( x right)>0$ and we have a convex function. Now Using Jensen’s inequality:
$$frac13fleft( fracA2 right)+frac13fleft( fracB2 right)+frac13fleft( fracC2 right)ge fleft( frac13fracA2+frac13fracB2+frac13fracC2 right)$$

share|cite|improve this answer

answered Jul 25 at 20:23

Vincent Law

1,457212

HINT:

Consider
$$fleft( x right)=1/sin left( fracx2 right);,quad 0<x<pi $$

It easy to verify that $f''left( x right)>0$ and we have a convex function. Now Using Jensen’s inequality:
$$frac13fleft( fracA2 right)+frac13fleft( fracB2 right)+frac13fleft( fracC2 right)ge fleft( frac13fracA2+frac13fracB2+frac13fracC2 right)$$

share|cite|improve this answer

answered Jul 25 at 20:23

Vincent Law

1,457212

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 20:23

Vincent Law

1,457212

answered Jul 25 at 20:23

Vincent Law

1,457212

answered Jul 25 at 20:23

Vincent Law

1,457212

1,457212

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks man this is simple after some show you the solution
  – Johnny chad
  Jul 25 at 20:37
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks man this is simple after some show you the solution
  – Johnny chad
  Jul 25 at 20:37
  

  
  
  
  

thanks man this is simple after some show you the solution
– Johnny chad
Jul 25 at 20:37

thanks man this is simple after some show you the solution
– Johnny chad
Jul 25 at 20:37

add a comment | 

up vote
3
down vote

A Pure Geometric Proof

Let $I$ be the incenter of $triangle ABC$, $r$ be the radius of the incircle. Then $$dfracrsindfracA2+dfracrsindfracB2+dfracrsindfracC2=IA+IB+IC.tag1$$

Now, let's apply Erdos-Mordell Inequality，which states that:

from a point $P$ inside a given $triangle ABC$， the perpendiculars $PX,
PY, PZ$ are drawn to its sides. Then

$$PA+PB+PC≥2(PX+PY+PZ).$$

Thus, we instantly have

$$IA+IB+ICgeq 2(r+r+r)=6r.tag2$$
Combining $(1)$ and $(2)$, we readily obtain what we want to prove.

share|cite|improve this answer

edited Jul 26 at 7:39

answered Jul 25 at 21:20

mengdie1982

2,835216

add a comment | 

up vote
3
down vote

A Pure Geometric Proof

Let $I$ be the incenter of $triangle ABC$, $r$ be the radius of the incircle. Then $$dfracrsindfracA2+dfracrsindfracB2+dfracrsindfracC2=IA+IB+IC.tag1$$

Now, let's apply Erdos-Mordell Inequality，which states that:

from a point $P$ inside a given $triangle ABC$， the perpendiculars $PX,
PY, PZ$ are drawn to its sides. Then

$$PA+PB+PC≥2(PX+PY+PZ).$$

Thus, we instantly have

$$IA+IB+ICgeq 2(r+r+r)=6r.tag2$$
Combining $(1)$ and $(2)$, we readily obtain what we want to prove.

share|cite|improve this answer

edited Jul 26 at 7:39

answered Jul 25 at 21:20

mengdie1982

2,835216

add a comment | 

up vote
3
down vote

up vote
3
down vote

A Pure Geometric Proof

Let $I$ be the incenter of $triangle ABC$, $r$ be the radius of the incircle. Then $$dfracrsindfracA2+dfracrsindfracB2+dfracrsindfracC2=IA+IB+IC.tag1$$

Now, let's apply Erdos-Mordell Inequality，which states that:

from a point $P$ inside a given $triangle ABC$， the perpendiculars $PX,
PY, PZ$ are drawn to its sides. Then

$$PA+PB+PC≥2(PX+PY+PZ).$$

Thus, we instantly have

$$IA+IB+ICgeq 2(r+r+r)=6r.tag2$$
Combining $(1)$ and $(2)$, we readily obtain what we want to prove.

share|cite|improve this answer

edited Jul 26 at 7:39

answered Jul 25 at 21:20

mengdie1982

2,835216

A Pure Geometric Proof

Let $I$ be the incenter of $triangle ABC$, $r$ be the radius of the incircle. Then $$dfracrsindfracA2+dfracrsindfracB2+dfracrsindfracC2=IA+IB+IC.tag1$$

Now, let's apply Erdos-Mordell Inequality，which states that:

from a point $P$ inside a given $triangle ABC$， the perpendiculars $PX,
PY, PZ$ are drawn to its sides. Then

$$PA+PB+PC≥2(PX+PY+PZ).$$

Thus, we instantly have

$$IA+IB+ICgeq 2(r+r+r)=6r.tag2$$
Combining $(1)$ and $(2)$, we readily obtain what we want to prove.

share|cite|improve this answer

edited Jul 26 at 7:39

answered Jul 25 at 21:20

mengdie1982

2,835216

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 7:39

edited Jul 26 at 7:39

edited Jul 26 at 7:39

answered Jul 25 at 21:20

mengdie1982

2,835216

answered Jul 25 at 21:20

mengdie1982

2,835216

answered Jul 25 at 21:20

mengdie1982

2,835216

2,835216

add a comment | 

add a comment | 

up vote
2
down vote

We have that

$$f(x)=frac1sin x$$

is convex then by Jensen's inequality

$$fracfrac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)3ge frac1sinleft(fracA+B+C6right)$$

and then

$$frac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)ge frac3sinleft(fracA+B+C6right)=6$$

share|cite|improve this answer

answered Jul 25 at 20:22

gimusi

65k73583

add a comment | 

up vote
2
down vote

We have that

$$f(x)=frac1sin x$$

is convex then by Jensen's inequality

$$fracfrac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)3ge frac1sinleft(fracA+B+C6right)$$

and then

$$frac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)ge frac3sinleft(fracA+B+C6right)=6$$

share|cite|improve this answer

answered Jul 25 at 20:22

gimusi

65k73583

add a comment | 

up vote
2
down vote

up vote
2
down vote

We have that

$$f(x)=frac1sin x$$

is convex then by Jensen's inequality

$$fracfrac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)3ge frac1sinleft(fracA+B+C6right)$$

and then

$$frac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)ge frac3sinleft(fracA+B+C6right)=6$$

share|cite|improve this answer

answered Jul 25 at 20:22

gimusi

65k73583

We have that

$$f(x)=frac1sin x$$

is convex then by Jensen's inequality

$$fracfrac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)3ge frac1sinleft(fracA+B+C6right)$$

and then

$$frac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)ge frac3sinleft(fracA+B+C6right)=6$$

share|cite|improve this answer

answered Jul 25 at 20:22

gimusi

65k73583

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 20:22

gimusi

65k73583

answered Jul 25 at 20:22

gimusi

65k73583

answered Jul 25 at 20:22

gimusi

65k73583

65k73583

add a comment | 

add a comment | 

up vote
0
down vote

Also, by AM-GM in the standard notation we obtain:
$$sum_cycfrac1sinfracalpha2geqfrac3sqrt[3]prodlimits_cycsinfracalpha2=frac3sqrt[3]fracr4Rgeqfrac3sqrt[3]frac14cdot2=6.$$

share|cite|improve this answer

answered Jul 25 at 20:40

Michael Rozenberg

87.8k1578180

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down vote?
  – Michael Rozenberg
  Jul 28 at 3:32
  

  
  
  
  

add a comment | 

up vote
0
down vote

Also, by AM-GM in the standard notation we obtain:
$$sum_cycfrac1sinfracalpha2geqfrac3sqrt[3]prodlimits_cycsinfracalpha2=frac3sqrt[3]fracr4Rgeqfrac3sqrt[3]frac14cdot2=6.$$

share|cite|improve this answer

answered Jul 25 at 20:40

Michael Rozenberg

87.8k1578180

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  Why someone down vote?
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Also, by AM-GM in the standard notation we obtain:
$$sum_cycfrac1sinfracalpha2geqfrac3sqrt[3]prodlimits_cycsinfracalpha2=frac3sqrt[3]fracr4Rgeqfrac3sqrt[3]frac14cdot2=6.$$

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answered Jul 25 at 20:40

Michael Rozenberg

87.8k1578180

Also, by AM-GM in the standard notation we obtain:
$$sum_cycfrac1sinfracalpha2geqfrac3sqrt[3]prodlimits_cycsinfracalpha2=frac3sqrt[3]fracr4Rgeqfrac3sqrt[3]frac14cdot2=6.$$

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answered Jul 25 at 20:40

Michael Rozenberg

87.8k1578180

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 20:40

Michael Rozenberg

87.8k1578180

answered Jul 25 at 20:40

Michael Rozenberg

87.8k1578180

answered Jul 25 at 20:40

Michael Rozenberg

87.8k1578180

87.8k1578180

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down vote?
  – Michael Rozenberg
  Jul 28 at 3:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why someone down vote?
  – Michael Rozenberg
  Jul 28 at 3:32
  

  
  
  
  

Why someone down vote?
– Michael Rozenberg
Jul 28 at 3:32

Why someone down vote?
– Michael Rozenberg
Jul 28 at 3:32

add a comment | 

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We use $$sinleft(fracA2right)=sqrtfrac(s-b)(s-c)bc$$ etc then we get by AM-GM:
$$frac13left(sqrtfracbc(s-b)(s-c)+sqrtfracac(s-a)(s-c)+sqrtfracab(s-a)(s-b)right)geq sqrt[3]fracabc(s-a)(s-b)(s-c)geq 2$$
if $$fracabc(s-a)(s-b)(s-c)geq 8$$

and this is $$abcgeq (-a+b+c)(a-b+c)(a+b-c)$$
using the Substitution

$$a=y+z,b=x+z,c=x+y$$

we get

$$(x+y)(x+z)(z+x)geq 8xyz$$ and this is AM-GM!
Note that $$s=fraca+b+c2$$

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answered Jul 25 at 20:45

Dr. Sonnhard Graubner

66.7k32659

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We use $$sinleft(fracA2right)=sqrtfrac(s-b)(s-c)bc$$ etc then we get by AM-GM:
$$frac13left(sqrtfracbc(s-b)(s-c)+sqrtfracac(s-a)(s-c)+sqrtfracab(s-a)(s-b)right)geq sqrt[3]fracabc(s-a)(s-b)(s-c)geq 2$$
if $$fracabc(s-a)(s-b)(s-c)geq 8$$

and this is $$abcgeq (-a+b+c)(a-b+c)(a+b-c)$$
using the Substitution

$$a=y+z,b=x+z,c=x+y$$

we get

$$(x+y)(x+z)(z+x)geq 8xyz$$ and this is AM-GM!
Note that $$s=fraca+b+c2$$

share|cite|improve this answer

answered Jul 25 at 20:45

Dr. Sonnhard Graubner

66.7k32659

add a comment | 

up vote
0
down vote

up vote
0
down vote

We use $$sinleft(fracA2right)=sqrtfrac(s-b)(s-c)bc$$ etc then we get by AM-GM:
$$frac13left(sqrtfracbc(s-b)(s-c)+sqrtfracac(s-a)(s-c)+sqrtfracab(s-a)(s-b)right)geq sqrt[3]fracabc(s-a)(s-b)(s-c)geq 2$$
if $$fracabc(s-a)(s-b)(s-c)geq 8$$

and this is $$abcgeq (-a+b+c)(a-b+c)(a+b-c)$$
using the Substitution

$$a=y+z,b=x+z,c=x+y$$

we get

$$(x+y)(x+z)(z+x)geq 8xyz$$ and this is AM-GM!
Note that $$s=fraca+b+c2$$

share|cite|improve this answer

answered Jul 25 at 20:45

Dr. Sonnhard Graubner

66.7k32659

We use $$sinleft(fracA2right)=sqrtfrac(s-b)(s-c)bc$$ etc then we get by AM-GM:
$$frac13left(sqrtfracbc(s-b)(s-c)+sqrtfracac(s-a)(s-c)+sqrtfracab(s-a)(s-b)right)geq sqrt[3]fracabc(s-a)(s-b)(s-c)geq 2$$
if $$fracabc(s-a)(s-b)(s-c)geq 8$$

and this is $$abcgeq (-a+b+c)(a-b+c)(a+b-c)$$
using the Substitution

$$a=y+z,b=x+z,c=x+y$$

we get

$$(x+y)(x+z)(z+x)geq 8xyz$$ and this is AM-GM!
Note that $$s=fraca+b+c2$$

share|cite|improve this answer

answered Jul 25 at 20:45

Dr. Sonnhard Graubner

66.7k32659

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 20:45

Dr. Sonnhard Graubner

66.7k32659

answered Jul 25 at 20:45

Dr. Sonnhard Graubner

66.7k32659

answered Jul 25 at 20:45

Dr. Sonnhard Graubner

66.7k32659

66.7k32659

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Applying Lagrange Multipliers shows that $$ sin(A/2) + sin (B/2) + sin (B/2)$$ is maximized when the triangle is equilateral in which case we have $$ sin(A/2) + sin (B/2) + sin (B/2)=3/2$$

The Arithmetic Harmonic inequality implies $$frac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)ge frac9sin(A/2) +sin(B/2)+sin(C/2)ge frac 9(3/2)=6$$

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answered Jul 25 at 20:47

Mohammad Riazi-Kermani

27.4k41852

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Applying Lagrange Multipliers shows that $$ sin(A/2) + sin (B/2) + sin (B/2)$$ is maximized when the triangle is equilateral in which case we have $$ sin(A/2) + sin (B/2) + sin (B/2)=3/2$$

The Arithmetic Harmonic inequality implies $$frac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)ge frac9sin(A/2) +sin(B/2)+sin(C/2)ge frac 9(3/2)=6$$

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answered Jul 25 at 20:47

Mohammad Riazi-Kermani

27.4k41852

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up vote
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down vote

up vote
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Applying Lagrange Multipliers shows that $$ sin(A/2) + sin (B/2) + sin (B/2)$$ is maximized when the triangle is equilateral in which case we have $$ sin(A/2) + sin (B/2) + sin (B/2)=3/2$$

The Arithmetic Harmonic inequality implies $$frac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)ge frac9sin(A/2) +sin(B/2)+sin(C/2)ge frac 9(3/2)=6$$

share|cite|improve this answer

answered Jul 25 at 20:47

Mohammad Riazi-Kermani

27.4k41852

Applying Lagrange Multipliers shows that $$ sin(A/2) + sin (B/2) + sin (B/2)$$ is maximized when the triangle is equilateral in which case we have $$ sin(A/2) + sin (B/2) + sin (B/2)=3/2$$

The Arithmetic Harmonic inequality implies $$frac1sin left( fracA2 right)+frac1sin left( fracB2 right)+frac1sin left( fracC2 right)ge frac9sin(A/2) +sin(B/2)+sin(C/2)ge frac 9(3/2)=6$$

share|cite|improve this answer

answered Jul 25 at 20:47

Mohammad Riazi-Kermani

27.4k41852

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 20:47

Mohammad Riazi-Kermani

27.4k41852

answered Jul 25 at 20:47

Mohammad Riazi-Kermani

27.4k41852

answered Jul 25 at 20:47

Mohammad Riazi-Kermani

27.4k41852

27.4k41852

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