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Consider three positive reals $x,y,z$ such that $xyz=1$.

How would one go about proving:

$\frac{x^5y^5}{x^2+y^2}+\frac{y^5z^5}{y^2+z^2}+\frac{x^5z^5}{x^2+z^2}\ge \frac{3}{2}$

I really dont know even where to begin! It looks a BIT like Nesbitts? Maybe?

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    @leon: I would think the question having both the [tag:inequality] and [tag:contest-math] tags would be adequate; what'd you have in mind?2011-09-03

5 Answers 5

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By Holder $\sum_{cyc}\frac{x^5y^5}{x^2+y^2}\geq\frac{(xy+xz+yz)^5}{(1+1+1)^3\sum\limits_{cyc}(x^2+y^2)}=\frac{(xy+xz+yz)^5}{54(x^2+y^2+z^2)}.$ Hence, it remains to prove that $(xy+xz+yz)^5\geq81(x^2+y^2+z^2).$ Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Thus, since $xy+xz+yz$ and $x^2+y^2+z^2$ does not depend on $w^3$,

it's enough to prove the last inequality for a maximal value of $w^3$, which happens for equality case of two variables.

Let $y=x$. Hence, $z=\frac{1}{x^2}$ and we need to prove that $(x^3+2)^5\geq81x(2x^6+1).$ Let $x^3=t$.

Hence, we need to prove that $f(t)\geq0$, where $f(t)=5\ln(t+2)-\ln(2t^2+1)-\frac{1}{3}\ln{t}-4\ln3$ But $f'(t)=\frac{5}{t+2}-\frac{4t}{2t^2+1}-\frac{1}{3t}=\frac{(t-1)(2t-1)(4t-1)}{3t(t+2)(2t^2+1)}$ and since $f\left(\frac{1}{4}\right)>0$, we are done!

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It actually can be reduced to Nesbitt's inequality quite quickly. Homogenize the equation by dividing it by $x^3y^3z^3 (= 1)$. Then what you want to prove is ${x^2y^2 \over x^2z^2 + y^2z^2} + {y^2z^2 \over x^2y^2 + x^2z^2} + {x^2z^2 \over x^2y^2 + y^2z^2} \geq {3 \over 2}$ This is exactly Nesbitt's inequality with parameters $x^2y^2$, $x^2z^2$, and $y^2z^2$.

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    Very nice. @Brian I guess you accepted an answer too soon :-). In any case, as a general suggestion, it may be good to wait for may be a few days time before accepting the most helpful answer.2011-09-02
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Edit. I posted a new "proof" (now deleted), before I realized I was adddressing the wrong question. I think the original proof is ok, I messed up something in trying to simplify the approach.

Hint.

Since this is an olympiad problem, it is likely that there is a proof without using calculus. I am sketching one here, but I have a feeling it can improved vastly.

Step 1. Make the substitution $x = 1/a$, $y = 1/b$ and $z = 1/c$, so that $abc = 1$. We are now left with the expression $ \sum \frac{1}{a^3b^3 (a^2+b^2)} = \sum \frac{c^3}{a^2+b^2} = \frac{a^3}{b^2+c^2} + \frac{b^3}{c^2+a^2} + \frac{c^3}{a^2+b^2}. $

Step 2. Assume the order $a \leq b \leq c$ without loss of generality. Then show that $ \frac{a^2}{b^2+c^2} \leq \frac{b^2}{c^2+a^2} \leq \frac{c^2}{a^2+b^2}. $ Now, by the "Chebyshev sum inequality", we have: $ \frac{a^3}{b^2+c^2} + \frac{b^3}{c^2+a^2} + \frac{c^3}{a^2+b^2} \geq \frac{1}{3} (a+b+c) \cdot \left( \frac{a^2}{b^2+c^2} + \frac{b^2}{c^2+a^2} + \frac{c^2}{a^2+b^2} \right). $

Step 3. For any $u,v,w > 0$, prove that $ \frac{u}{v+w} + \frac{v}{w+u} + \frac{w}{u+v} \geq \frac{3}{2}. $

Step 4. Conclude the inequality by plugging in the third inequality in the second.

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    I just learned that the inequality in my Step 3 is called Nesbitt's inequality ([wikipedia link](http://en.wikipedia.org/wiki/Nesbitt's_inequality)). Cool!2011-09-02
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Sorry, I don't know latex. But, my solution is the following:

  1. Get rid of denominators.

  2. Multiply by 2.

  3. Expand de products. You'll get cyclic sums.

  4. Left side will be something like: cyc(7, 7, 0) + cyc(7, 5, 2) + cyc(5, 5, 4)

  5. Homogenize right side by multiplicating by cubic root of xyz to the 8th power. That way the sums of exponents of both sides of this inequality will be 14. There will be 24 terms on the left side and 24 terms on the right side.

  6. Finally, you have to prove:

2( cyc(7, 7, 0) + cyc(7, 5, 2) + cyc(5, 5, 4) ) ≥ 3 ( cyc(20/3, 14/3, 8/3) + 2*cyc(14/3, 14/3, 14/3)

  1. But, that happens to be Muirhead's Theorem. Done!
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    this site may help with LaTex: http://co$d$ecogs.com/latex/eqne$d$itor.php2011-10-03
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Look at this pic. or solution. Hope you got a simple standard approach...

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    Could you include the work in the body of the post instead of linking to a picture? Not only is it bothersome to click out to a link but it can be pretty hard to read.2017-10-07