Ta có:[tex]\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2bc}}+\frac{1}{2\sqrt{ab ^2c}}+\frac{1}{2\sqrt{abc^2}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}[/tex]
Ta có:[tex]\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2bc}}+\frac{1}{2\sqrt{ab ^2c}}+\frac{1}{2\sqrt{abc^2}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}[/tex]