Giải thích các bước giải:
Ta có :
$b^2+c^2\ge 2bc$
$\begin{split}\rightarrow \dfrac{bc}{a^2+2bc}&=\dfrac{1}{2}.\dfrac{2bc}{a^2+2bc}\\&=\dfrac{1}{2}. (1-\dfrac{a^2}{a^2+2bc})\\&\le\dfrac{1}{2}.(1-\dfrac{a^2}{a^2+b^2+c^2})\end{split}$
Tương tự ta có :
$\dfrac{ca}{b^2+2ca}\le\dfrac{1}{2}.(1-\dfrac{b^2}{a^2+b^2+c^2})$
$\dfrac{ab}{c^2+2ab}\le\dfrac{1}{2}.(1-\dfrac{c^2}{a^2+b^2+c^2})$
$\rightarrow \dfrac{bc}{a^2+2bc}+\dfrac{ca}{b^2+2ca}+\dfrac{ab}{c^2+2ab}\le 1$