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$\frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2}\le 1$$\Leftrightarrow \frac{2}{a^2+2}+\frac{2}{b^2+2}+\frac{2}{c^2+2}\le 2$$\Leftrightarrow \frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}\ge 1$Áp dụng BĐT Bunhia$\frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}$$\ge \frac{(a+b+c)^2}{a^2+b^2+c^2+6}$$=\frac{a^2+b^2+c^2+2ab+2bc+2ca}{a^2+b^2+c^2}=1$

$\frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2}\le 1$$\Leftrightarrow \frac{2}{a^2+2}+\frac{2}{b^2+2}+\frac{2}{c^2+2}\le 2$$\Leftrightarrow \frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}\ge 1$Áp dụng BĐT Bunhia$\frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}$$\ge \frac{(a+b+c)^2}{a^2+b^2+c^2+6}$$=\frac{a^2+b^2+c^2+2ab+2bc+2ca}{a^2+b^2+c^2+6}=1$