Question

[Nguyễn Thị Mát]

Cho a ; b >0 và \(a^2+b^2=1\) . Tìm GTNN của bieeyr thức :

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

Answer 1

[Kudo Shinichi]

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 \(a=b=\frac{1}{\sqrt{2}}\)

Answer 2

[Nyatmax]

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 \(a=b=\frac{1}{\sqrt{2}}\)