Cho a, b, c > 0; $\sum a^{2}=1.$Chứng minh rằng:
$\sum \frac{a^{2}+1}{b^{2}+ac}\geq 6$
Nguồn: Sáng tác.
Ta có : $\sum \frac{a^{2}+1}{b^{2}+ac}\geq \sum \frac{a^{2}+1}{b^{2}+\frac{1}{2}(a^{2}+c^{2})}=\sum \frac{2(a^{2}+1)}{2b^{2}+a^{2}+c^{2}}=\frac{2(a^{2}+1)}{b^{2}+1}+\frac{2(b^{2}+1)}{c^{2}+1}+\frac{2(c^{2}+1)}{a^{2}+1}\geq 3\sqrt[3]{\frac{2^{3}(a^{2}+1)(b^{2}+1)(c^{2}+1)}{(c^{2}+1)(b^{2}+1)(a^{2}+1)}}=6.$