Prove $\sum\limits_{cyc}\sqrt[3]{\frac{a^2+bc}{b+c}}\ge\sqrt[3]{9(a+b+c)}$ Let $a,b,c$ be positive real numbers.  Show that
$$\sum_{cyc}\sqrt[3]{\dfrac{a^2+bc}{b+c}}\ge\sqrt[3]{9(a+b+c)}$$
I have tried C-S and  Holder inequalities, without success.  How to solve it? 

Just a little hint 
By AM GM
$$\sum \sqrt[3] {\frac {a^2+bc}{b+c}}\ge 3\sqrt[3] {\prod \left (\frac {a^2}{b+c}+ \frac {bc}{b+c}\right)^{\frac {1}{3}}}$$
Now by Hölder's inequality 
$$3\sqrt[3] {\prod \left(\frac {a^2}{b+c}+ \frac {bc}{b+c}\right)^{\frac {1}{3}}}\ge 3\sqrt[3] {\left (\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}\right)^{\frac {1}{3}}+\left (\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}\right)^{\frac {1}{3}}}=3\sqrt[3] {2\sqrt[3] {\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}}}$$
Now again by AM GM on denominator 
$$3\sqrt[3] {2\sqrt[3] {\frac {a^2b^2c^2}{(a+b)(b+c)(a+c)}}}\ge 3\sqrt[3] {\frac {3}{(a+b+c)}\sqrt[3] {a^2b^2c^2}}$$
Hope it helped any way
