Question

[QUan]

Cho \(a\ge1,b\ge1\)

Cm: \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)

Answer 1

[Hoàng Lê Bảo Ngọc]

Chứng minh bằng biến đổi tương đương : 

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

\(\Leftrightarrow\left(\frac{1}{1+a^2}-\frac{1}{1+ab}\right)+\left(\frac{1}{1+b^2}-\frac{1}{1+ab}\right)\ge0\)

\(\Leftrightarrow\frac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\frac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow\left(\frac{a-b}{1+ab}\right)\left(\frac{b}{1+b^2}-\frac{a}{1+a^2}\right)\ge0\)

\(\Leftrightarrow\frac{a-b}{1+ab}.\frac{\left(a-b\right)\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(ab+1\right)\left(a^2+1\right)\left(b^2+1\right)}\ge0\)

Vì \(a\ge1,b\ge1\) nên \(ab-1\ge0\) . Mặt khác vì \(\left(a-b\right)^2\ge0\) nên ta có điều phải chứng minh.

Answer 2

[huyhieurong75]

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)