ta có : \(a+b=1\Rightarrow\left\{{}\begin{matrix}2a+b=1+a\\a+2b=1+b\end{matrix}\right.\)
ta có : \(M=\left(1+\dfrac{1}{a}\right)^2+\left(1+\dfrac{1}{b}\right)^2\)
\(\Leftrightarrow M=\left(\dfrac{1+a}{a}\right)^2+\left(\dfrac{1+b}{b}\right)^2\)
\(\Leftrightarrow M=\left(\dfrac{2a+b}{a}\right)^2+\left(\dfrac{2b+a}{b}\right)^2\)
\(\Leftrightarrow M=\left(2+\dfrac{b}{a}\right)^2+\left(2+\dfrac{a}{b}\right)^2\)
\(\Leftrightarrow M=4+\dfrac{4b}{a}+\dfrac{b^2}{a^2}+4+\dfrac{4a}{b}+\dfrac{a^2}{b^2}\)
ta có : \(\dfrac{4b}{a}+\dfrac{4a}{b}\ge2\sqrt{\dfrac{4b}{a}.\dfrac{4a}{b}}=2\sqrt{16}=8\) ( côsi )
và \(\dfrac{b^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{b^2}{a^2}.\dfrac{a^2}{b^2}}=2\sqrt{1}=2\) (côsi )
\(\Rightarrow M=4+\dfrac{4b}{a}+\dfrac{b^2}{a^2}+4+\dfrac{4a}{b}+\dfrac{a^2}{b^2}\ge18\)dấu " = " xảy ra \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\dfrac{4b}{a}=\dfrac{4a}{b}\\\dfrac{b^2}{a^2}=\dfrac{a^2}{b^2}\end{matrix}\right.\Leftrightarrow a=b=0,5\)
vậy giá trị nhỏ nhất của \(M\) là \(18\) khi \(a=b=0,5\)
