Question

Cho a,b > 0 và \(a^2+b^2=1\). Chứng minh : \(\left(1+a\right)\left(a+\frac{1}{b}\right)+\left(1+b\right)\left(b+\frac{1}{a}\right)\ge3\left(1+\sqrt{2}\right)\)

Answer 1

Áp dụng BĐT AM-GM ta có: 

\(VT=a^2+b^2+\frac{a}{b}+\frac{b}{a}+\frac{1}{a}+\frac{1}{b}+a+b\)

\(=1+\frac{a}{b}+\frac{b}{a}+\frac{1}{a}+\frac{1}{b}+a+b\)

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

\(\ge3+2\sqrt{\frac{1}{a}\cdot2a}+2\sqrt{\frac{1}{b}\cdot2b}-\sqrt{2\left(a^2+b^2\right)}\)

\(\ge3+4\sqrt{2}-\sqrt{2}=3+3\sqrt{2}=3\left(1+\sqrt{2}\right)\)

Khi \(a=b=\frac{1}{\sqrt{2}}\)