# work 019 – C : 2790 Gazeta Mathematica

For $a,b,c > 0$, prove that

$\frac {a^3}{2a + 3b} + \frac {b^3}{2b + 3c} + \frac {c^3}{2c + 3a}\geq \frac {a^2 + b^2 + c^2}{5}$

Proof

$\sum\frac{a^{3}}{2a+3b}=\frac{a^{4}}{2a^{2}+3ab}\ge\sum\frac{(a^{2}+b^{2}+c^{2})^{2}}{2(a^{2}+b^{+}c^{2})+3(ab+bc+ca)}\ge\frac{(a^{2}+b^{2}+c^{2})^{2}}{5(a^{2}+b^{2}+c^{2})}=LHS$

I have used Cauchy-Schwartz Inequality.