Áp dụng BĐT Cauchy:
$\frac{1}{2}(\frac{1}{ab}+\frac{1}{ac})\geq \frac{1}{a\sqrt{bc}} \geq \frac{2}{a^{2}+bc}$
$\frac{1}{2}(\frac{1}{bc}+\frac{1}{ba})\geq \frac{1}{b\sqrt{ca}} \geq\frac{2}{b^{2}+ca}$
$\frac{1}{2}(\frac{1}{ca}+\frac{1}{cb})\geq \frac{1}{c\sqrt{ab}} \geq \frac{2}{c^{2}+ab}$
$\Rightarrow\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ab}\leq\frac{1}{2}(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})=\frac{a+b+c}{2abc}$
Dấu "=" xảy ra $\Leftrightarrow a=b=c$