Inequality

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4 posts

#1 h Apr 15, 2018, 6:20 PM • 2 Y

Y by Adventure10, Mango247

If $a, b, c$ are positive real numbers such that $abc=1$ , then prove the following:

$\frac{1}{a^6 + b^6 +1} + \frac{1}{b^6 + c^6 +1} + \frac{1}{c^6 + a^6 +1} \leq 1$

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#2 h Apr 15, 2018, 6:25 PM • 2 Y

Y by Adventure10, Mango247

By C-S $(a^6+b^6+1)(\frac{1}{a^2}+\frac{1}{b^2}+c^4)\ge(\sum a^2)^2$ $\implies$ we should prove $\frac{2\sum a^2b^2+\sum a^4}{(\sum a^2)^2}\le1$ which is obvious

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#3 h Apr 15, 2018, 6:41 PM • 2 Y

Y by Adventure10, Mango247

Actually the last inequality is always an equality.

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561 posts

#4 h Apr 15, 2018, 7:04 PM • 2 Y

Y by Adventure10, Mango247

I find the substitution $x=a^6$ very useful; it reduces the problem to an easier case!

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#5 h Apr 15, 2018, 7:11 PM • 1 Y

Y by Adventure10

achen29 wrote:

I find the substitution $x=a^6$ very useful; it reduces the problem to an easier case!

How do you solve it after the substitutions?

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486 posts

#6 h Apr 15, 2018, 7:12 PM • 2 Y

Y by Adventure10, Mango247

better is $x=a^2$ then use $x^3+y^3 \geq xy(x+y)$

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547 posts

#7 h Apr 15, 2018, 7:13 PM • 2 Y

Y by Adventure10, Mango247

Alternate solution using denotions
$a^6=\frac{1}{x}$ ... gives $\frac{1}{a^6+b^6+1}=\frac{1}{\frac{1}{x}+\frac{1}{y}+1}=\frac{xy}{x+y+xy}=\frac{xyz}{xz+yz+xyz}$ then again by C-S $(xz+yz+1)(xz+yz+x^2y^2)\ge(\sum xy)^2$ which follows we should prove $\frac{2\sum xy+\sum x^2y^2}{(\sum xy)^2}\le1$ where we have equality so done

This post has been edited 1 time. Last edited by georgeado17, Apr 15, 2018, 7:19 PM
Reason: typo

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#8 h Apr 15, 2018, 7:15 PM • 2 Y

Y by Adventure10, Mango247

georgeado17 wrote:

$\frac{xy}{x+y+xy}=\frac{xyz}{xz+yz+xy}$

check this !

This post has been edited 1 time. Last edited by Muradjl, Apr 15, 2018, 7:15 PM

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#9 h Apr 15, 2018, 7:26 PM • 2 Y

Y by Adventure10, Mango247

We could also use another substitutions like $a^6=\frac{x}{y}$ ...anyway it follows to use C-S
These substitutions give that $\frac{1}{a^6+b^6+1}=\frac{1}{\frac{x}{y}+\frac{y}{z}+1}$ and by C-S $(\frac{x}{y}+\frac{y}{z}+1)(xy+yz+z^2)\ge(\sum x)^2$ $\implies$ need to prove that $\frac{2\sum xy+\sum x^2}{(\sum x)^2}\le1$ where again we have equality and again it's done!

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#10 h Apr 15, 2018, 7:36 PM • 2 Y

Y by Adventure10, Mango247

MuradjI's idea is also perfect cause such kind of a substitution gives $\sum\frac{z}{x+y+z}\le1$ which is obvious and which doesn't use C-S but also there exists solution using C-S and cute just denote $a^3=x$ ...it gives
$\frac{1}{a^6+b^6+1}=\frac{1}{x^2+y^2+1}\le\frac{z^2+2}{(x+y+z)^2}$ whis one follows that we need to prove $xy+yz+xz\ge3$ which is obvious by AM-GM in case $xyz=1$

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547 posts

#11 h Apr 15, 2018, 7:41 PM • 2 Y

Y by Adventure10, Mango247

In last provement it's not required to denote $a^3$ finally we will get $\sum a^3b^3\ge3$ which is obvious sorry

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561 posts

#12 h Apr 15, 2018, 7:58 PM • 2 Y

Y by Adventure10, Mango247

Muradjl wrote:

better is $x=a^2$ then use $x^3+y^3 \geq xy(x+y)$

lol I came to this conclusion but is it too late to saaaaay change the substitution ?

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tr2512

#13 h Apr 16, 2018, 12:07 PM • 2 Y

Y by Adventure10, Mango247

USAMO 1997

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#14 h Apr 16, 2018, 2:15 PM • 2 Y

Y by Adventure10, Mango247

georgeado17 wrote:

we should prove $\frac{2\sum a^2b^2+\sum a^4}{(\sum a^2)^2}\le1$ which is obvious

Can anyone explain how to proceed from here?

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#15 h Apr 16, 2018, 3:23 PM • 2 Y

Y by Adventure10, Mango247

Evenprime123 wrote:

georgeado17 wrote:

we should prove $\frac{2\sum a^2b^2+\sum a^4}{(\sum a^2)^2}\le1$ which is obvious

Can anyone explain how to proceed from here?

WolfusA wrote:

Actually the last inequality is always an equality.

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