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After experimenting, I've come to the conclusion that if $x\geq y\geq z\geq 0:$

$$\sum_{x,y,z}\frac{x}{\sqrt{x+y}}\geq \sum_{x,y,z}\frac{y}{\sqrt{x+y}}$$

(the sums are cyclic)

Does anyone know how to prove this? I've tried variants of Cauchy Schwarz, and the rearrangement inequality, but these are not powerful enough.

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  • $\begingroup$ Do you consider a cyclic sum? Or the sum is all over possible combinations? $\endgroup$ – mfl Jul 2 '14 at 12:49
  • $\begingroup$ @mfl Cyclic sum $\endgroup$ – user161248 Jul 2 '14 at 12:51
  • $\begingroup$ @YvesDaoust and $\frac{z}{\sqrt{z+x}}\leq \frac{x}{\sqrt{z+x}}\dots$ $\endgroup$ – user161248 Jul 2 '14 at 13:28
  • $\begingroup$ @YvesDaoust That is exactly what I wrote in my post. In the comment I was pointing at that your suggestion does not help. $\endgroup$ – user161248 Jul 2 '14 at 13:46
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The condition $x^2+y^2+z^2=1$ is redundant.

Let $a=\sqrt{y+z},b=\sqrt{z+x},c=\sqrt{x+y}$ we have $c\ge b \ge a \ge 0$ and $$x=\frac{b^2 + c^2 - a^2}{2},y=\frac{c^2 + a^2 - b^2}{2},z=\frac{a^2 + b^2 - c^2}{2}.$$ Replacing into the original inequality, it becomes $$\sum_{a,b,c} \frac{b^2 + c^2 - a^2}{2c} \ge \sum_{a,b,c} \frac{c^2 + a^2 - b^2}{2c},$$ which is true because \begin{align} LHS-RHS &= \sum_{a,b,c} \frac{b^2 - a^2}{c} \\ &= \frac{b^2 - a^2}{c} + \frac{c^2 - b^2}{a} +\frac{(a^2-b^2) +(b^2 - c^2)}{b}\\ &= (b^2 - a^2)\left(\frac{1}{c}-\frac{1}{b}\right) + (c^2 - b^2)\left(\frac{1}{a}-\frac{1}{b}\right) \\ &= \frac{(b-a)(c-b)}{b} \left(-\frac{b+a}{c}+\frac{b+c}{a}\right) \\ &= \frac{(b-a)(c-b)(c-a)(a+b+c)}{abc}\\ &\ge 0. \end{align}

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  • $\begingroup$ Thanks, I've taken the condition out. $\endgroup$ – user161248 Jul 2 '14 at 13:35
  • $\begingroup$ You're welcome ;) $\endgroup$ – Khue Jul 2 '14 at 13:42
  • $\begingroup$ @user161248 I think that a condition $x\ge y \ge z >0$ is needed. Otherwise, when $x=y=z=0$, the solution above is not applicable. $\endgroup$ – mike Jul 2 '14 at 14:45

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