Theo đề bài thì: \(ab+bc+ca=3abc\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
\(\sum\dfrac{a}{a^2+bc}\le\sum\dfrac{a}{2a\sqrt{bc}}=\sum\dfrac{1}{2\sqrt{bc}}\)
\(\le\dfrac{1}{2}\sum\left(\dfrac{1}{2a}+\dfrac{1}{2b}\right)=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{3}{2}\)