\(M=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2\)
Thay \(a+b=1\) vào biểu thức:
\(\Rightarrow M=\left(1+\frac{a+b}{a}\right)^2+\left(1+\frac{a+b}{b}\right)^2\)
\(M=\left(2+\frac{b}{a}\right)^2+\left(2+\frac{a}{b}\right)^2\)
Áp dụng BĐT Cô - si:
\(\Rightarrow\left\{\begin{matrix}\left(2+\frac{b}{a}\right)^2\ge\frac{8b}{a}\\\left(2+\frac{a}{b}\right)^2\ge\frac{8a}{b}\end{matrix}\right.\)
\(\Rightarrow M=\left(2+\frac{b}{a}\right)^2+\left(2+\frac{a}{b}\right)^2\ge\frac{8b}{a}+\frac{8a}{b}=8\left(\frac{b}{a}+\frac{a}{b}\right)\)
Áp dụng BĐT Cô - si:
\(\Rightarrow\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{ab}{ab}}=2\)
\(\Rightarrow8\left(\frac{b}{a}+\frac{a}{b}\right)\ge16\)
Mà \(M=\left(2+\frac{b}{a}\right)^2+\left(2+\frac{a}{b}\right)^2\ge8\left(\frac{b}{a}+\frac{a}{b}\right)\)
\(\Rightarrow M=\left(2+\frac{b}{a}\right)^2+\left(2+\frac{a}{b}\right)^2\ge16\)
\(\Leftrightarrow M=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2\ge16\)
Vậy GTNN của \(M=16\)
BĐt phụ: \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\)và \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)(cauchy)
cm: áp dụng BĐT bunhiacopxki:\(\left(1+1\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\Leftrightarrow a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\)(dấu = xảy ra khi a=b)
áp dụng vào bài toán ta có:
\(M=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2\ge\frac{1}{2}\left(2+\frac{1}{a}+\frac{1}{b}\right)^2\ge\frac{1}{2}\left(2+\frac{4}{a+b}\right)^2\)
\(M\ge\frac{1}{2}\left(2+4\right)^2=18\)
dấu =xảy ra khi a=b=0,5
vậy M min =18 khi a=b=1/2
