For $a,b,c$. Prove that $\frac{a^2}{a+\sqrt[3]{bc}}+\frac{b^2}{b+\sqrt[3]{ca}}+\frac{c^2}{c+\sqrt[3]{ab}}\ge\frac{3}{2}$

Question

For $a,b,c$ are positive real number satisfy $a+b+c=3$. Prove that $$\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{3}{2}$$

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By Cauchy-Schwarz: $\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}$

Hence we need prove $\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\dfrac{3}{2} (*)$

$\Leftrightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}$

By AM-GM $a+b+1\geq 3\sqrt[3]{ab}$ and $...$

$\Rightarrow2\left(a+b+c\right)+3\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}$

$\Rightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}$

I need a new method.

Answer 1

By AM-GM and C-S we obtain: $$\sum_{cyc}\frac{a^2}{a+\sqrt[3]{bc}}\geq\sum_{cyc}\frac{a^2}{a+\frac{b+c+1}{3}}=\sum_{cyc}\frac{9a^2}{3(3a+b+c)+a+b+c}=$$ $$=\sum_{cyc}\frac{9a^2}{10a+4b+4c}\geq\frac{9(a+b+c)^2}{\sum\limits_{cyc}(10a+4b+4c)}=\frac{9(a+b+c)^2}{18(a+b+c)}=\frac{3}{2}$$