Given positive real numbers $a,b,c$ satisfying $a^2+b^2+c^2=9.$ Prove that$$\color{black}{\frac{1}{\sqrt[3]{a^3+b^3}}+\frac{1}{\sqrt[3]{c^3+b^3}}+\frac{1}{\sqrt[3]{a^3+c^3}}\ge \sqrt[3]{\frac{a+b+c}{2}}. }$$
By $a=b=c=\sqrt{3}$ the inequality become an equality.
Thus, we need to prove that
For all $a,b,c>0$ then $$\color{blue}{\frac{1}{\sqrt[3]{a^3+b^3}}+\frac{1}{\sqrt[3]{c^3+b^3}}+\frac{1}{\sqrt[3]{a^3+c^3}}\ge \sqrt[3]{\frac{81(a+b+c)}{2(a^2+b^2+c^2)^2}}. }$$
A big problem we have around the point $c=\dfrac{\sqrt{538}}{59}a=\dfrac{\sqrt{538}}{59}b$ because the difference between the LHS and the RHS in this point is $0.000975...$
I think the problem is ver sharp and naive AM-GM, C-S is hard to apply.
Could you given some hint to prove it? Thank you.
Some thoughts.
By Holder inequality, we have \begin{align*} &\left(\sum_{\mathrm{cyc}} \frac{1}{\sqrt[3]{a^3+b^3}}\right)^3 \left(\sum_{\mathrm{cyc}} (a^3 + b^3)(a + b + 2c)^4 \right)\\[6pt] \ge{}& \left(\sum_{\mathrm{cyc}} (a + b + 2c)\right)^4. \end{align*}
It suffices to prove that $$\left(\sum_{\mathrm{cyc}} (a + b + 2c)\right)^4 \ge \frac{81(a+b+c)}{2(a^2+b^2+c^2)^2}\sum_{\mathrm{cyc}} (a^3 + b^3)(a + b + 2c)^4. \tag{1}$$
(1) is true which is verified by Mathematica.