\(P=\left(b+c+d\right)\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)=1+\frac{b}{c}+\frac{b}{d}+\frac{c}{b}+1+\frac{c}{d}+\frac{d}{b}+\frac{d}{c}+1\)
\(=3+\frac{b}{c}+\frac{b}{d}+\frac{c}{d}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}\)
Mặt khác do \(b\le c\le d\Rightarrow\left(d-c\right)\left(c-b\right)\ge0\)
\(\Leftrightarrow cd-bd-c^2+bc\ge0\Leftrightarrow bc+cd\ge c^2+bd\)
\(\Leftrightarrow\frac{bc+cd}{cd}\ge\frac{c^2+bd}{cd}\Leftrightarrow\frac{b}{d}+1\ge\frac{c}{d}+\frac{b}{c}\)
\(\frac{bc+cd}{bc}\ge\frac{c^2+bd}{bc}\Leftrightarrow\frac{d}{b}+1\ge\frac{c}{b}+\frac{d}{c}\)
\(\Leftrightarrow\frac{b}{d}+\frac{d}{b}+2\ge\frac{b}{c}+\frac{c}{d}+\frac{c}{b}+\frac{d}{c}\)
\(\Leftrightarrow2\left(\frac{b}{d}+\frac{d}{b}\right)+2\ge\frac{b}{c}+\frac{b}{d}+\frac{c}{d}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}=P\)
Mà \(a\le b\le d\le2a\Rightarrow\left\{{}\begin{matrix}\frac{1}{2}\le\frac{b}{d}\le1\\1\le\frac{d}{b}\le2\end{matrix}\right.\)
\(\Rightarrow\left(\frac{b}{d}-1\right)\left(\frac{d}{b}-2\right)\ge0\Leftrightarrow1-2\frac{b}{d}-\frac{d}{b}+2\ge0\)
\(\Leftrightarrow\frac{b}{d}+\frac{d}{b}\le3-\frac{b}{d}\le3-\frac{1}{2}=\frac{5}{2}\)
\(\Rightarrow P\le2.\frac{5}{2}+2=7\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}b=c=a\\d=2a\end{matrix}\right.\)
