Áp dụng BĐT AM - GM
\(A=\left(a+1\right)\left(1+\frac{1}{b}\right)+\left(b+1\right)\left(1+\frac{1}{a}\right)\)
\(=\frac{a}{b}+\frac{b}{a}+a+\frac{1}{a}+b+\frac{1}{b}+2\)
\(=\frac{a}{b}+\frac{b}{a}+\left(a+\frac{1}{2a}\right)+\left(b+\frac{1}{2b}\right)+\frac{1}{2a}+\frac{1}{2b}+2\)
\(\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{a.\frac{1}{2a}}+2\sqrt{b.\frac{1}{2b}}+2\sqrt{\frac{1}{2a}.\frac{1}{2b}}+2\)
\(=4+2\sqrt{2}+\frac{1}{\sqrt{ab}}\ge4+2\sqrt{2}+\frac{1}{\frac{\sqrt{2\left(a^2+b^2\right)}}{2}}\)
\(=4+3\sqrt{2}\)
Dấu " = " xảy ra khi \(a=b=\frac{1}{\sqrt{2}}\)
Ta co:\(1=a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\Rightarrow a+b\le\sqrt{2}\)
Ta lai co:
\(A=\frac{a}{b}+\frac{b}{a}+\frac{1}{a}+\frac{1}{b}+a+b+2\)
\(=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{1}{a}+2a\right)+\left(\frac{1}{b}+2b\right)-\left(a+b\right)+2\)
\(\ge2+2\sqrt{2}+2\sqrt{2}-\sqrt{2}+2=4+3\sqrt{2}\)
Dau '=' xay ra khi \(a=b=\frac{1}{\sqrt{2}}\)
Vay \(A_{min}=4+3\sqrt{2}\)khi \(a=b=\frac{1}{\sqrt{2}}\)
Bài làm:
Ta có: \(A=\left(1+a\right)\left(1+\frac{1}{b}\right)+\left(1+b\right)\left(1+\frac{1}{a}\right)\)
\(=1+a+\frac{1}{b}+\frac{a}{b}+1+b+\frac{1}{a}+\frac{b}{a}\)
\(=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(a+b\right)+\left(\frac{1}{a}+\frac{1}{b}\right)+2\)
\(\ge2.\sqrt{\frac{a}{b}.\frac{b}{a}}+2.\sqrt{ab}+2.\sqrt{\frac{1}{a}.\frac{1}{b}}\)
\(=2+2\sqrt{ab}+\frac{2}{\sqrt{ab}}\)
\(\ge2+2.\sqrt{2\sqrt{ab}.\frac{2}{\sqrt{ab}}}\)
\(=2+2.2=6\)
Dấu "=" xảy ra khi: \(a=b=\frac{1}{\sqrt{2}}\)
Vậy \(Min_A=6\Leftrightarrow a=b=\frac{1}{\sqrt{2}}\)