# $\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$

Prove : $$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$$ For $a,b,c,d>0$

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Why do you ask? What have you tried? – Did Aug 23 '11 at 11:54
It should be solvable using bunching. en.wikipedia.org/wiki/Muirhead%27s_inequality#Examples – David Schwartz Aug 23 '11 at 12:49
@David: if I wanted to do that, the denominators in the LHS and the lack of symmetry of both sides would annoy me. But maybe you know how to circumvent these... – Did Aug 23 '11 at 13:02
I'm tring to solve it use holder inequality ،then i've got : $$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left‌​(\frac{a+d}{2}\right)^3+\left(\frac{b+c}{2}\right)^3$$ ..is it usefull ? – Lina Aug 23 '11 at 13:07
@Lina: Kudos for showing your work. This reasoning could succeed if $(a+d)^3+(b+c)^3\ge(a+b)^3+(c+d)^3$ but there is no reason to believe this last inequality holds (and it happens to be false in general). – Did Aug 23 '11 at 13:19

I know many time has passed, however i came up with a solution and I hope it is correct: here you are:

Let us denote $$\begin{array}{c}\frac{1}{4}\frac{(b+c)^3}{2}(8(a^2+d^2)^3-(a+d)^3(a+b)^3)+\\\frac{1}{4}\frac{(a+d)^3}{2}(8(b^2+c^2)^3-(c+d)^3(c+b)^3)=(\clubsuit)\end{array}$$

And we want to prove $(\clubsuit)\geq0.$

We recall $x^3-y^3=(x-y)(x^2+xy+y^2)$ $(\spadesuit)$ and so $$\begin{array}{c}(\clubsuit)=(2(a^2+d^2)-(a+d)(a+b))C_1+2((b^2+c^2)-(c+d)(c+b))C_2\geq\\ \min\{C_1,C_2\}\cdot(2(a^2+b^2+c^2+d^2)-(a+d)(a+b)-(c+d)(c+d)).\end{array}$$

Where $C_1,C_2$ are nonnegative costants depending on $a,b,c,d$, and can be easily derived following $(\spadesuit)$

But now it is true that $$2(a^2+b^2+c^2+d^2)-a^2-ab-ad-bd-c^2-cd-cb-bd\geq 0,$$ since $$\begin{eqnarray} b^2+d^2&\geq& 2bd,\\\frac{a^2+d^2}{2}&\geq& ad,\\ \frac{c^2+d^2}{2}&\geq& cd,\\\frac{c^2+b^2}{2}&\geq&cb,\\\frac{a^2+b^2}{2}&\geq&ab,\\a^2&\geq&a^2,\\c^2&\geq&c^2.\end{eqnarray}$$

We have then estabilished $(\clubsuit)\geq 0$. But it is easy to show that this relation implies $$(a^2+d^2)^3(b+c)^3+(b^2+c^2)^3(a+d)^3\geq\frac{(a+b)^3(a+d)^3(b+c)^3}{8}+\frac{(c+d)^3(a+d)^3(b+c)^3}{8}$$ which in turn implies $$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq \left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3,$$ as desired.

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Proof by Texas Hold'em :) – user13838 Sep 15 '11 at 15:14
$\clubsuit$, $\heartsuit$, $\spadesuit$, $\diamondsuit$ are the best symbols in Latex. They combine two of my great passions: maths and Hold'em :) – uforoboa Sep 15 '11 at 15:53
Thank you very much – Lina Sep 16 '11 at 14:17