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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    $\begingroup$ You might start by noting that you get equality when $x=y=z=1$. You also have from the arithmetic-geometric mean inequality that $x+y+z \ge 3$, and probably much more. $\endgroup$ – Henry Sep 1 '11 at 22:43
  • $\begingroup$ Use lagrange-multipliers to minimize the lhs and get the desired result. $\endgroup$ – Listing Sep 1 '11 at 22:46
  • $\begingroup$ Also note that this is really similar to this question: math.stackexchange.com/questions/57057/… $\endgroup$ – Listing Sep 1 '11 at 22:47
  • $\begingroup$ Lagrange multipliers leads to a veryyy bashy system of equations. To prove that x=y=z would require major algebra skills. Also, i dont see how the question resembles the one in your link at all...? $\endgroup$ – Brian Sep 1 '11 at 23:01
  • $\begingroup$ Inequalities/extrema involving symmetric rational functions are so common (specially in olympiads)... I wonder if a tag should be added. $\endgroup$ – leonbloy Sep 2 '11 at 18:28

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.


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.

  • $\begingroup$ I was about to type pretty much the same, so everything seems ok to me ;) $\endgroup$ – leshik Sep 1 '11 at 23:36
  • $\begingroup$ I just learned that the inequality in my Step 3 is called Nesbitt's inequality (wikipedia link). Cool! $\endgroup$ – Srivatsan Sep 2 '11 at 12:56

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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    $\begingroup$ 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. $\endgroup$ – Srivatsan Sep 2 '11 at 14:50

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!

Look at this pic. or solution. Hope you got a simple standard approach...

  • $\begingroup$ 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. $\endgroup$ – Sriotchilism O'Zaic Oct 7 '17 at 14:40

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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