Giải thích các bước giải:

Ta có:

$\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}-\dfrac{2}{1+ab}$

$=(\dfrac{1}{1+a^2}-\dfrac{1}{1+ab})+(\dfrac{1}{1+b^2}-\dfrac{1}{1+ab})$

$=\dfrac{1+ab-(1+a^2)}{(1+a^2)(1+ab)}+\dfrac{1+ab-(1+b^2)}{(1+b^2)(1+ab)}$

$=\dfrac{a(b-a)}{(1+a^2)(1+ab)}+\dfrac{b(a-b)}{(1+b^2)(1+ab)}$

$=\dfrac{a-b}{1+ab}\cdot (\dfrac{-a}{1+a^2}+\dfrac{b}{1+b^2})$

$=\dfrac{a-b}{1+ab}\cdot \dfrac{-a(1+b^2)+b(1+a^2)}{(1+a^2)(1+b^2)}$

$=\dfrac{a-b}{1+ab}\cdot \dfrac{-a-ab^2+b+ba^2}{(1+a^2)(1+b^2)}$

$=\dfrac{a-b}{1+ab}\cdot \dfrac{-(a-b)-ab(b-a)}{(1+a^2)(1+b^2)}$

$=\dfrac{a-b}{1+ab}\cdot \dfrac{-(a-b)+ab(a-b)}{(1+a^2)(1+b^2)}$

$=\dfrac{a-b}{1+ab}\cdot \dfrac{(a-b)(ab-1)}{(1+a^2)(1+b^2)}$

$= \dfrac{(a-b)^2(ab-1)}{(1+a^2)(1+b^2)(1+ab)}$

Mà $a\ge b\ge 1\to ab-1\ge 0$

$\to \dfrac{(a-b)^2(ab-1)}{(1+a^2)(1+b^2)(1+ab)}\ge 0$

$\to \dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}-\dfrac{2}{1+ab}\ge 0$

$\to \dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge \dfrac{2}{1+ab}$

$\to đpcm$