Question

Cm: với \(ab\ge1\) thì \(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\)

Answer 1

Ta có :

\(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\)⇒\(\left(\dfrac{1}{1+a^2}-\dfrac{1}{1+ab}\right)+\left(\dfrac{1}{1+b^2}-\dfrac{1}{1+ab}\right)\ge0\)

\(\dfrac{ab-a^2}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{ab-b^2}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)\(\dfrac{\left(b-a\right)^2\left(ab-1\right)}{\left(1+b^2\right)\left(1+a^2\right)\left(1+ab\right)}\ge0\).vì ab≥0 nên sua ra đpcm