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 $a,b,c,d>0$
$\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$
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1@David: putting aside tediousness considerations, I mainly wonder about the non symmetry (which should invalidate this approach, unless I am mistaken). – 2011-08-23
2 Answers
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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1$\clubsuit$, $\heartsuit$, $\spadesuit$, $\diamondsuit$ are the best symbols in Latex. They combine two of my great passions: maths and Hold'em :) – 2011-09-15
Since for $x>0$ we have $\left(\frac{x^2+1}{x+1}\right)^3-\frac{x^3+1}{2}=\frac{(x-1)^4(x^2+x+1)}{2(x+1)^3}\geq0$ and by P-M $\left(\frac{x+1}{2}\right)^3\leq\frac{x^3+1}{2},$ we obtain: $\left(\frac{a^2+d^2}{a+d}\right)^3+\left(\frac{b^2+c^2}{b+c}\right)^3\geq\frac{a^3+d^3}{2}+\frac{b^3+c^3}{2}=$ $=\frac{a^3+b^3}{2}+\frac{c^3+d^3}{2}\geq\left(\frac{a+b}{2}\right)^3+\left(\frac{c+d}{2}\right)^3$ and we are done!
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1@Hans It's$a$privet case of Mildorf's inequality. For $k\geq1$ and positives $a$ and $b$ we have $\frac{(1+k)(a-b)^2+8ab}{4(a+b)}\geq\left(\frac{a^k+b^k}{2}\right)^{\frac{1}{k}}$ – 2017-03-12