An inequality in four variables Let $a$, $b$, $c$ and $d$ be positive real numbers. Prove that: 
$$\frac{ab+bc+ca}{a^3+b^3+c^3}+\frac{ab+bd+da}{a^3+b^3+d^3}+\frac{ac+cd+da}{a^3+c^3+d^3}+\frac{bc+cd+db}{b^3+c^3+d^3}\le\min\left [\frac{a^2+b^2}{(ab)^{\frac 32}}+\frac{c^2+d^2}{(cd)^{\frac 32}},\frac{a^2+c^2}{(ac)^{\frac 32}}+\frac{b^2+d^2}{(bd)^{\frac 32}},\frac{a^2+d^2}{(ad)^{\frac 32}}+\frac{b^2+c^2}{(bc)^{\frac 32}}\right ].$$
My Solution as follows:
$\frac{ab+bc+ca}{a^3+b^3+c^3}\le \frac{ab+bc+ca}{3abc}=\frac 13\left(\frac 1a+\frac 1b+\frac 1c\right)$
$\implies\frac{ab+bc+ca}{a^3+b^3+c^3}+\frac{ab+bd+da}{a^3+b^3+d^3}+\frac{ac+cd+da}{a^3+c^3+d^3}+\frac{bc+cd+db}{b^3+c^3+d^3}\le\left(\frac 1a+\frac 1b+\frac 1c+\frac 1d\right)$
And 
$a^2+b^2\ge a^{\frac32}b^{\frac12}+a^{\frac12}b^{\frac32}$ (By Rearrangement)
$\implies \frac{a^2+b^2}{(ab)^{\frac 32}}\ge\frac{a+b}{ab}=\frac 1a+\frac 1b$
$\implies \frac{a^2+b^2}{(ab)^{\frac 32}}+\frac{c^2+d^2}{(cd)^{\frac 32}}\ge\frac1a+\frac1b+\frac1c+\frac1d$
Similarly each term of Right hand side is greater than or equal to $\frac1a+\frac1b+\frac1c+\frac1d$
Hence the inequality.
 A: We need to prove that
$$\sum_{cyc}\left(\frac{ab+ac+bc}{a^3+b^3+c^3}-\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\right)\leq$$
$$\leq\min\left \{\tfrac{a^2+b^2}{(ab)^{\frac 32}}-\tfrac{1}{a}-\tfrac{1}{b}+\tfrac{c^2+d^2}{(cd)^{\frac 32}}-\tfrac{1}{c}-\tfrac{1}{d},\tfrac{a^2+c^2}{(ac)^{\frac 32}}-\tfrac{1}{a}-\tfrac{1}{c}+\tfrac{b^2+d^2}{(bd)^{\frac 32}}-\tfrac{1}{b}-\tfrac{1}{d},\tfrac{a^2+d^2}{(ad)^{\frac 32}}-\tfrac{1}{a}-\tfrac{1}{d}+\tfrac{b^2+c^2}{(bc)^{\frac 32}}-\tfrac{1}{b}-\tfrac{1}{c}\right \}$$ or
$$\sum_{cyc}\frac{(ab+ac+bc)(a^3+b^3+c^3-3abc)}{3abc(a^3+b^3+c^3)}+$$
$$+\min\left\{\tfrac{a^2+b^2-(a+b)\sqrt{ab}}{\sqrt{a^3b^3}}+\tfrac{c^2+d^2-(c+d)\sqrt{cd}}{\sqrt{c^3d^3}},\tfrac{a^2+c^2-(a+c)\sqrt{ac}}{\sqrt{a^3c^3}}+\tfrac{b^2+d^2-(b+d)\sqrt{bd}}{\sqrt{b^3d^3}},\tfrac{a^2+d^2-(a+d)\sqrt{ad}}{\sqrt{a^3d^3}}+\tfrac{b^2+c^2-(b+c)\sqrt{bc}}{\sqrt{b^3c^3}}\right\}\geq0,$$
which is true by Muirhead because $(3,0,0)\succ(1,1,1)$ and $(2,0)\succ\left(\frac{3}{2},\frac{1}{2}\right)$.
Done! 
