Mình có cách này, mọi người c0i có được không.
$Q.e.D\Leftrightarrow \sum \left ( \frac{a^{2}}{b+c}-\frac{a}{2} \right )\geq \frac{3\sum a^{3}-\sum a^{2}\sum a}{2\sum a^{2}}$
$\Leftrightarrow \frac{a\left ( a-b \right )+a\left ( a-c \right )}{b+c}+\frac{b\left ( b-a \right )+b\left ( b-c \right )}{a+c}+\frac{c\left ( c-a \right )+c\left ( c-b \right )}{a+b}\geqslant \frac{\sum a^{2}\left [ \left ( a-b \right )+\left ( a-c \right ) \right ]}{a^{2}+b^{2}+c^{2}}$
$\Leftrightarrow \sum \left ( a-b \right )\left ( \frac{a}{b+c}-\frac{b}{c+a} \right )\geq \frac{\sum \left ( a-b \right )\left ( a^{2}-b^{2} \right )}{a^{2}+b^{2}+c^{2}}$
$\Leftrightarrow \sum \left ( a-b \right )\frac{\left ( a^{2}-b^{2} \right )+c\left ( a-b \right )}{\left ( a+c \right )\left ( b+c \right )}\geq \frac{\sum \left ( a-b \right )\left ( a^{2}-b^{2} \right )}{a^{2}+b^{2} +c^{2}}$
$\Leftrightarrow \sum \left ( a-b \right )^{2}\left [ \frac{a+b+c}{\left ( a+c \right )\left ( b+c \right )}-\frac{a+b}{a^{2}+b^{2}+c^{2}} \right ]\geq 0$
Bất đẳng thức trên luôn đúng do:
$\left ( a+b+c \right )\left ( a^{2}+b^{2}+c^{2} \right )> \frac{\left [ \left ( a+b \right )+\left ( b+c \right )+\left ( c+a \right ) \right ]^{3}}{27}\geq \left ( a+b \right )\left ( b+c \right )\left ( c+a \right )\square$