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Find the minimum of $(1+a^2)(1+b^2)(1+c^2)$ where $a,b,cgeq 0$

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3

$begingroup$

Find the minimum of:

$$(1+a^2)(1+b^2)(1+c^2) a,b,cgeq 0$$

Knowing that $$ab+bc+ac=27$$

I tried my best using QM-AM-GM inequalities, Cauchy-Schwarz, etc. I tried also to do it with partial derivatives but it's too long. Thank you for your time :)

inequality contest-math

share|cite|improve this question

edited 12 hours ago

YuiTo Cheng

2,3184937

asked 13 hours ago

EurekaEureka

712113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My initial thought: Let $x=ab,y=bc,z=ac$, so that the constraint function is $x+y+z=27$. Then $xy=ab^2 c=b^2 zimplies b^2=x y/z$ and similarly $a^2=xz/y$, $c^2=yz/x$. That makes the objective function more annoying, but it still seems tractable.
  $endgroup$
  – Semiclassical
  12 hours ago
  
  

  
  
  
  


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  $begingroup$
  Can you provide a link to the contest? Or some other evidence that the contest is not currently active. I have no reason to think that it is, but we have this rule :-)
  $endgroup$
  – Jyrki Lahtonen
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JyrkiLahtonen it's just a simulation of math team national italian olympiad. It's not even an official competition :) .The true Olympiads are in May as you can see here olimpiadi.dm.unibo.it (but the site is italian).
  $endgroup$
  – Eureka
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, Eureka.
  $endgroup$
  – Jyrki Lahtonen
  7 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Find the minimum of:

$$(1+a^2)(1+b^2)(1+c^2) a,b,cgeq 0$$

Knowing that $$ab+bc+ac=27$$

I tried my best using QM-AM-GM inequalities, Cauchy-Schwarz, etc. I tried also to do it with partial derivatives but it's too long. Thank you for your time :)

inequality contest-math

share|cite|improve this question

edited 12 hours ago

YuiTo Cheng

2,3184937

asked 13 hours ago

EurekaEureka

712113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My initial thought: Let $x=ab,y=bc,z=ac$, so that the constraint function is $x+y+z=27$. Then $xy=ab^2 c=b^2 zimplies b^2=x y/z$ and similarly $a^2=xz/y$, $c^2=yz/x$. That makes the objective function more annoying, but it still seems tractable.
  $endgroup$
  – Semiclassical
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you provide a link to the contest? Or some other evidence that the contest is not currently active. I have no reason to think that it is, but we have this rule :-)
  $endgroup$
  – Jyrki Lahtonen
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JyrkiLahtonen it's just a simulation of math team national italian olympiad. It's not even an official competition :) .The true Olympiads are in May as you can see here olimpiadi.dm.unibo.it (but the site is italian).
  $endgroup$
  – Eureka
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, Eureka.
  $endgroup$
  – Jyrki Lahtonen
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Find the minimum of:

$$(1+a^2)(1+b^2)(1+c^2) a,b,cgeq 0$$

Knowing that $$ab+bc+ac=27$$

I tried my best using QM-AM-GM inequalities, Cauchy-Schwarz, etc. I tried also to do it with partial derivatives but it's too long. Thank you for your time :)

inequality contest-math

share|cite|improve this question

edited 12 hours ago

YuiTo Cheng

2,3184937

asked 13 hours ago

EurekaEureka

712113

$endgroup$

share|cite|improve this question

edited 12 hours ago

YuiTo Cheng

2,3184937

asked 13 hours ago

EurekaEureka

712113

share|cite|improve this question

edited 12 hours ago

YuiTo Cheng

2,3184937

asked 13 hours ago

EurekaEureka

712113

edited 12 hours ago

YuiTo Cheng

2,3184937

edited 12 hours ago

YuiTo Cheng

2,3184937

asked 13 hours ago

EurekaEureka

712113

asked 13 hours ago

EurekaEureka

712113

2 Answers
2

active

oldest

votes

13

$begingroup$

We note that $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$, and $c$. We rewrite it by using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$:
$$beginalign
(1+a^2)(1+b^2)(1+c^2)&=(ab+bc+ca-1)^2+(a+b+c-abc)^2\
&=26^2+(a+b+c-abc)^2geq 26^2=676
endalign$$
where the condition $ab+bc+ac=27$ has been applied.

Hence the minimum is $676$ as soon as we show that there are $a,b,cgeq 0$ such that $ab+bc+ac=27$ and $a+b+c=abc$.
Can you take it from here?

share|cite|improve this answer

edited 12 hours ago

answered 12 hours ago

Robert ZRobert Z

101k1072145

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Robert Z Amazing! +1
  $endgroup$
  – Michael Rozenberg
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How had you this idea ? +1
  $endgroup$
  – Jean Marie
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$ and $c$. Try to rewrite it using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$.
  $endgroup$
  – Robert Z
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  I have a rather direct proof of the identity used by @Robert Z : use twice what is called Diophantes equality $(a^2+b^2)(c^2+d^2)=(ac−bd)^2+(ad+bc)^2$ (which can be explained as two ways of writing the squared norm ( = squared module) of $(a+ib)(c+id)$)... Or, said in an even simpler way take the squared norm of both sides of relationship $(1+ia)(1+ib)(1+ic)=(1-ab-bc-ca)+(a+b+c-abc)i$.
  $endgroup$
  – Jean Marie
  11 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Good job. You even managed to convince Michael about the usefulness of elementary symmetric polynomials - something I failed the last time I tried.
  $endgroup$
  – Jyrki Lahtonen
  11 hours ago
  

  
  
  
  

 | 
show 5 more comments

2

$begingroup$

A simple supplement to Robert's nice answer. With the aid of the following plot I'm sure you all can prove that when $a+b+c=12=abc$ the polynomial
$$(x-a)(x-b)(x-c)=x^3-12x^2+27x-12$$
has three positive zeros $a,b,c$ :-)

share|cite|improve this answer

answered 11 hours ago

community wiki

Jyrki Lahtonen

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Adding this because it was not immediately obvious to me how to prove that Robert's lower bound can be achieved. I'm sure there are nice trick ways of doing that. But, it really is simple.
  $endgroup$
  – Jyrki Lahtonen
  11 hours ago
  

  
  
  
  

add a comment | 

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13

$begingroup$

We note that $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$, and $c$. We rewrite it by using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$:
$$beginalign
(1+a^2)(1+b^2)(1+c^2)&=(ab+bc+ca-1)^2+(a+b+c-abc)^2\
&=26^2+(a+b+c-abc)^2geq 26^2=676
endalign$$
where the condition $ab+bc+ac=27$ has been applied.

Hence the minimum is $676$ as soon as we show that there are $a,b,cgeq 0$ such that $ab+bc+ac=27$ and $a+b+c=abc$.
Can you take it from here?

share|cite|improve this answer

edited 12 hours ago

answered 12 hours ago

Robert ZRobert Z

101k1072145

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Robert Z Amazing! +1
  $endgroup$
  – Michael Rozenberg
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How had you this idea ? +1
  $endgroup$
  – Jean Marie
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$ and $c$. Try to rewrite it using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$.
  $endgroup$
  – Robert Z
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  I have a rather direct proof of the identity used by @Robert Z : use twice what is called Diophantes equality $(a^2+b^2)(c^2+d^2)=(ac−bd)^2+(ad+bc)^2$ (which can be explained as two ways of writing the squared norm ( = squared module) of $(a+ib)(c+id)$)... Or, said in an even simpler way take the squared norm of both sides of relationship $(1+ia)(1+ib)(1+ic)=(1-ab-bc-ca)+(a+b+c-abc)i$.
  $endgroup$
  – Jean Marie
  11 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Good job. You even managed to convince Michael about the usefulness of elementary symmetric polynomials - something I failed the last time I tried.
  $endgroup$
  – Jyrki Lahtonen
  11 hours ago
  

  
  
  
  

 | 
show 5 more comments

13

$begingroup$

We note that $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$, and $c$. We rewrite it by using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$:
$$beginalign
(1+a^2)(1+b^2)(1+c^2)&=(ab+bc+ca-1)^2+(a+b+c-abc)^2\
&=26^2+(a+b+c-abc)^2geq 26^2=676
endalign$$
where the condition $ab+bc+ac=27$ has been applied.

Hence the minimum is $676$ as soon as we show that there are $a,b,cgeq 0$ such that $ab+bc+ac=27$ and $a+b+c=abc$.
Can you take it from here?

share|cite|improve this answer

edited 12 hours ago

answered 12 hours ago

Robert ZRobert Z

101k1072145

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Robert Z Amazing! +1
  $endgroup$
  – Michael Rozenberg
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How had you this idea ? +1
  $endgroup$
  – Jean Marie
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$ and $c$. Try to rewrite it using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$.
  $endgroup$
  – Robert Z
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  I have a rather direct proof of the identity used by @Robert Z : use twice what is called Diophantes equality $(a^2+b^2)(c^2+d^2)=(ac−bd)^2+(ad+bc)^2$ (which can be explained as two ways of writing the squared norm ( = squared module) of $(a+ib)(c+id)$)... Or, said in an even simpler way take the squared norm of both sides of relationship $(1+ia)(1+ib)(1+ic)=(1-ab-bc-ca)+(a+b+c-abc)i$.
  $endgroup$
  – Jean Marie
  11 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Good job. You even managed to convince Michael about the usefulness of elementary symmetric polynomials - something I failed the last time I tried.
  $endgroup$
  – Jyrki Lahtonen
  11 hours ago
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

We note that $(1+a^2)(1+b^2)(1+c^2)$ is a symmetric function with respect to $a$, $b$, and $c$. We rewrite it by using the elementary symmetric polynomials $a+b+c$, $ab+bc+ca$ and $abc$:
$$beginalign
(1+a^2)(1+b^2)(1+c^2)&=(ab+bc+ca-1)^2+(a+b+c-abc)^2\
&=26^2+(a+b+c-abc)^2geq 26^2=676
endalign$$
where the condition $ab+bc+ac=27$ has been applied.

Hence the minimum is $676$ as soon as we show that there are $a,b,cgeq 0$ such that $ab+bc+ac=27$ and $a+b+c=abc$.
Can you take it from here?

share|cite|improve this answer

edited 12 hours ago

answered 12 hours ago

Robert ZRobert Z

101k1072145

$endgroup$

share|cite|improve this answer

edited 12 hours ago

answered 12 hours ago

Robert ZRobert Z

101k1072145

answered 12 hours ago

Robert ZRobert Z

101k1072145

answered 12 hours ago

Robert ZRobert Z

101k1072145