Question

Cho a.b \(\ge0\) và \(a^2+b^2=1\). Tìm GTNN của biểu thức:

A = \(\left(1+a\right)\left(1+\dfrac{1}{b}\right)+\left(1+b\right)\left(1+\dfrac{1}{a}\right)\)

Answer 1

Sửa: \(a;b>0\)

Áp dụng BĐT AM-GM ta có:

\(A=\left(a+1\right)\left(1+\dfrac{1}{b}\right)+\left(b+1\right)\left(1+\dfrac{1}{a}\right)\)

\(=\dfrac{a}{b}+\dfrac{b}{a}+a+\dfrac{1}{a}+b+\dfrac{1}{b}+2\)

\(=\dfrac{a}{b}+\dfrac{b}{a}+\left(a+\dfrac{1}{2a}\right)+\left(b+\dfrac{1}{2b}\right)+\dfrac{1}{2a}+\dfrac{1}{2b}+2\)

\(\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}+2\sqrt{a\cdot\dfrac{1}{2a}}+2\sqrt{b\cdot\dfrac{1}{2b}}+2\sqrt{\dfrac{1}{2a}\cdot\dfrac{1}{2b}}+2\)

\(=4+2\sqrt{2}+\dfrac{1}{\sqrt{ab}}\)\(\ge4+2\sqrt{2}+\dfrac{1}{\dfrac{\sqrt{2\left(a^2+b^2\right)}}{2}}\)

\(=4+3\sqrt{2}\)

Dấu \("="\) xảy ra khi \(a=b=\dfrac{1}{\sqrt{2}}\)