Ta có: \(\left\{{}\begin{matrix}\dfrac{a}{bc\left(a+1\right)}=\dfrac{a}{abc+bc}=\dfrac{a}{\left(ab+bc\right)+\left(bc+ca\right)}\\\dfrac{b}{ca\left(b+1\right)}=\dfrac{b}{abc+ca}=\dfrac{b}{\left(ab+ca\right)+\left(bc+ca\right)}\\\dfrac{c}{ab\left(c+1\right)}=\dfrac{c}{abc+ab}=\dfrac{c}{\left(ab+bc\right)+\left(ca+ab\right)}\end{matrix}\right.\)(ab+bc+ca=abc)
Áp dụng bđt Svácxơ, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
\(P\le\dfrac{a}{4}\left[\dfrac{1}{bc+ca}+\dfrac{1}{ab+bc}\right]+\dfrac{b}{4}\left[\dfrac{1}{ab+ca}+\dfrac{1}{bc+ca}\right]+\dfrac{c}{4}\left[\dfrac{1}{ab+bc}+\dfrac{1}{ca+ab}\right]\)
\(=\dfrac{a}{4\left(a+b\right)c}+\dfrac{a}{4b\left(c+a\right)}+\dfrac{b}{4\left(b+c\right)a}+\dfrac{b}{4\left(a+b\right)c}+\dfrac{c}{4\left(c+a\right)b}+\dfrac{c}{4\left(b+c\right)a}\)
\(=\dfrac{a+b}{4\left(a+b\right)c}+\dfrac{b+c}{4\left(b+c\right)a}+\dfrac{c+a}{4\left(c+a\right)b}\)
\(=\dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{4c}=\dfrac{ab+bc+ca}{4abc}=\dfrac{abc}{4abc}=\dfrac{1}{4}\)(ab+bc+ca=abc)
\(MaxP=\dfrac{1}{4}\Leftrightarrow a=b=c=3\)
