Question

[Đỗ Hồng Ngọc]

Cho \(a\ge1,b\ge1.\)Chứng minh rằng \(\frac{1+ab}{1+a^2}+\frac{1+ab}{1+b^2}\ge2\)

Answer 1

[trần gia bảo]

Ta có: \(\frac{1+ab}{1+a^2}+\frac{1+ab}{1+b^2}=\left(1+ab\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}\right)\)

mà \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{4}{2+a^2+b^2}\)( Áp dụng BĐT phụ \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\))

Mặt khác: \(a^2+b^2\ge2ab\)

=> \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{4}{2+2ab}=\frac{2}{1+ab}\)

=> \(\left(1+ab\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}\right)\ge\left(1+ab\right)\left(\frac{2}{1+ab}\right)=2\)(đpcm)