Question

Cho a≥1 b≥1 thỏa mãn: a≥1 b≥1:

CMR

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

Answer 1

BĐT \(\Leftrightarrow\left(b^2+1\right)\left(ab+1\right)+\left(a^2+1\right)\left(ab+1\right)\ge2\left(a^2+1\right)\left(b^2+1\right)\)

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

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

\(\Leftrightarrow\left(a-b\right)\left(a^2b+b-a^2b-a\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(ab-1\right)\ge0\)(luôn đúng)