Ta có:$(a^{2}+2)(1+\frac{(b+c)^{2}}{2})\geq (a+b+c)^{2}\rightarrow \frac{1}{a^{2}+2}\leq \frac{1+\frac{(b+c)^{2}}{2}}{(a+b+c)^{2}}$.Tương tự rồi cộng vế vs vế ta có:
$\frac{1}{a^{2}+2}+\frac{1}{b^{2}+2}+\frac{1}{c^{2}+2}\leq \frac{3+a^{2}+b^{2}+c^{2}+ab+bc+ca}{(a+b+c)^{2}}$
$\leftrightarrow \frac{3+a^{2}+b^{2}+c^{2}+ab+bc+ca}{(a+b+c)^{2}}\geq 1\leftrightarrow ab+bc+ca\leq 3$