Prove: $\sqrt[3]{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\frac{c^{2}+ab}{a^2+b^2}}\geq 9\frac{\sqrt[3]{abc}}{a+b+c}$ Let $a,b,c\in \mathbb{R^+}$.

Prove: $\sqrt[3]{\frac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\frac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\frac{c^{2}+ab}{a^2+b^2}}\geq 9\frac{\sqrt[3]{abc}}{a+b+c}$

PS: I don't have ay ideas about this problem :(
Thanks
 A: By Holder $\left(\sum\limits_{cyc}\sqrt[3]{\frac{a^2+bc}{b^2+c^2}}\right)^3\sum\limits_{cyc}(b^2+c^2)(a^2+bc)^3\geq\left(\sum\limits_{cyc}(a^2+ab)\right)^4$.
Thus, it remains to prove that $\left(\sum\limits_{cyc}(a^2+ab)\right)^4(a+b+c)^3\geq729abc\sum\limits_{cyc}(b^2+c^2)(a^2+bc)^3$, which is obvious.
A: $\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}+\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}+\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}\ge 3 \sqrt[3]{\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}}  \\ \iff \sqrt[3]{\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}} \ge 3\dfrac{\sqrt[3]{abc}}{a+b+c} $
now we prove a stronger one :
$\sqrt[3]{\dfrac{a^{2}+bc}{b^2+c^2}}\sqrt[3]{\dfrac{b^{2}+ac}{a^2+c^2}}\sqrt[3]{\dfrac{c^{2}+ab}{a^2+b^2}}\ge \dfrac{3\sqrt[3]{abc}}{a+b+c} \iff \dfrac{(a^{2}+bc)(b^{2}+ac)(c^{2}+ab)}{(b^2+c^2)(a^2+c^2)(a^2+b^2)} \ge \dfrac{3^3abc}{(a+b+c)^3} \iff \dfrac{(a+b+c)^3}{3^3abc} \ge  \dfrac{(b^2+c^2)(a^2+c^2)(a^2+b^2)}{(a^{2}+bc)(b^{2}+ac)(c^{2}+ab)}$
there is a lemma:
$\dfrac{(a+b)(b+c)(c+a)}{8abc} \ge  \dfrac{(b^2+c^2)(a^2+c^2)(a^2+b^2)}{(a^{2}+bc)(b^{2}+ac)(c^{2}+ab)}$
Michael Rozenberg give this lemma a nice proof:
$(c^{2}+ab)(a+b)=a(b^2+c^2)+b(a^2+c^2) \ge 2 \sqrt{ab(b^2+c^2)(a^2+c^2)}$
with AM-GM, it is easy to prove:
$\dfrac{(a+b+c)^3}{3^3abc} \ge \dfrac{(a+b)(b+c)(c+a)}{8abc}$
A: This is just a hunch. First, the terms are all homogenous of the same degree. That's a good start. Put $a=b=c=1$. This gives exact equality. That's probably useful for something. Chances are, that's the maximum or the minimum of the difference between left and right hand side. Now vary $a$ by reducing it. Take the derivative. You can assume initially that $b=c=1$. Reduce $a$ from 1 to 0. Then the inequality gets stronger by the time you get to $a=0$. My hunch is that it's probably monotonic all the way from $a=1$ down to $a=0$. You can do the same sort of thing with the pair $(b,c)$, showing convexity or concavity. Then the result should magically appear! That's just a hunch.
PS. This is what I think the graph looks like.

That might give a few clues!
