40) Cho $a;b;c>0$ thỏa $a^2+b^2+c^2=1$. Cmr: $\sum \frac{a}{b^2+c^2}\geq \frac{3\sqrt{3}}{2}$
từ đk ta được BDT:
$\sum \frac{a}{1-a^2}\geq \frac{3\sqrt{3}}{2}$
ta có: $2=2\left ( 1-a^2 \right )+2a^2\geq 3\sqrt[3]{a^2(1-a^2)^2}\Rightarrow a(1-a^2)\geq \frac{4}{27}\Rightarrow \frac{a}{1-a^2}\geq \frac{3\sqrt{3}}{2}a^2\Rightarrow \sum \frac{a}{1-a^2}\geq \frac{3\sqrt{3}}{2}(\sum a^2)=\frac{3\sqrt{3}}{2}.; "="\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}$