Áp dụng BĐT: \(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\) ( Câu hỏi của ZoZ - Kudo vs Conan - ZoZ - Toán lớp 9 | Học trực tuyến)
\(\Rightarrow\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Áp dụng vào, ta có:
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{9}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)\)\(\dfrac{bc}{b+3c+2a}=\dfrac{bc}{\left(a+c\right)+\left(a+b\right)+2c}\le\dfrac{9}{bc}\left(\dfrac{1}{a+c}+\dfrac{1}{a+b}+\dfrac{1}{2c}\right)\)\(\dfrac{ca}{c+3a+2b}=\dfrac{ca}{\left(c+b\right)+\left(b+a\right)+2a}\le\dfrac{ca}{9}\left(\dfrac{1}{c+b}+\dfrac{1}{b+a}+\dfrac{1}{2a}\right)\)
Cộng vế theo vế BĐT, ta được:
\(P\le\dfrac{1}{9}\left(\dfrac{bc+ac}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ab+ac}{b+c}\right)+\dfrac{1}{18}\left(a+b+c\right)\)
\(P\le\dfrac{1}{9}\left[\dfrac{c\left(a+b\right)}{a+b}+\dfrac{b\left(c+a\right)}{a+c}+\dfrac{a\left(b+c\right)}{b+c}\right]+\dfrac{1}{18}\left(a+b+c\right)\)
\(P\le\dfrac{1}{9}\left(a+b+c\right)+\dfrac{1}{18}\left(a+b+c\right)\)
\(P\le\dfrac{1}{6}\left(a+b+c\right)\) \(=\dfrac{1}{6}.6=1\)
\(\Rightarrow Max_P=1\Leftrightarrow a=b=c=2\)
