46) Cho $a;b;c>0$ thỏa $a^2+b^2+c^2=1$. Cmr: $\sum \frac{1}{a^2+b^2}\leq \frac{\sum a^3}{2abc}+3$
$\sum \frac{1}{a^2+b^2}\leq \frac{\sum a}{2abc}\leq \frac{\sum a^3+6abc}{2abc}\leq \frac{\sum a^3}{2abc}+3."="\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}$