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

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.

$\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:

$\dfrac{(a+b+c)^3}{3^3abc} \ge \dfrac{(a+b)(b+c)(c+a)}{8abc}$