Ta cần đi chứng minh : [tex]\frac{1}{a^2+2} + \frac{1}{b^2+2} +\frac{1}{c^2+2}\leq 1 => \frac{\frac{a^2}{2}}{a^2+2} + \frac{\frac{b^2}{2}}{b^2+2} + \frac{\frac{c^2}{2}}{c^2+2} \geq \frac{1}{2}
=> \frac{a^2}{2(a^2+2)} + \frac{a^2}{2(a^2+2)} + \frac{c^2}{2(c^2+2)} \geq \frac{(a+b+c)^2}{2(a^2+b^2+c^2) + 12}= \frac{(a+b+c)^2}{2(a^2+b^2+c^2+2ab+2bc+2ac)} = \frac{(a+b+c)^2}{2(a+b+c)^2} = \frac{1}{2}[/tex]
Mình không chắc lắm:v