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