Plaginicj

2022-09-04

For $a,b,c$. Prove that $\frac{{a}^{2}}{a+\sqrt[3]{bc}}+\frac{{b}^{2}}{b+\sqrt[3]{ca}}+\frac{{c}^{2}}{c+\sqrt[3]{ab}}\ge \frac{3}{2}$
By Cauchy-Schwarz: $⇒\frac{{a}^{2}}{a+\sqrt[3]{bc}}+\frac{{b}^{2}}{b+\sqrt[3]{ca}}+\frac{{c}^{2}}{c+\sqrt[3]{ab}}\ge \frac{{\left(a+b+c\right)}^{2}}{a+b+c+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}$
Hence we need prove $\frac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge \frac{3}{2}\left(\ast \right)$
$⇔9\ge 3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}$
By AM-GM $a+b+1\ge 3\sqrt[3]{ab}$ and $...$
$⇒2\left(a+b+c\right)+3\ge 3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}$
$⇒9\ge 3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}$
I need a new method.

Belen Solomon

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

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