[不等式] new Vasc inequality

If $a,b,c\ge -3$ such that $a+b+c=3$, then
\[\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}\ge \frac 1{a}+\frac 1{b}+\frac 1{c}.\]