Ta có:
Xét hiệu $\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1} - \dfrac{2}{ab+1}$
$=\dfrac{a^2+b^2+2}{(a^2+1)(b^2+1)}-\dfrac{2}{ab+1}$
$=\dfrac{(a^2+b^2+2)(ab+1)-2(a^2+1)(b^2+1)}{(a^2+1)(b^2+1)(ab+1)}$
$=\dfrac{a^3b+b^3a+2ab+a^2+b^2+2-2(ab)^2-a^2-b^2-2}{(a^2+1)(b^2+1)(ab+1)}$
$=\dfrac{ab(a^2+b^2)+2ab-2(ab)^2-2}{(a^2+1)(b^2+1)(ab+1)}$
$=\dfrac{ab(a^2-2ab+b^2)+2(ab-1)}{(a^2+1)(b^2+1)(ab+1)}$
$=\dfrac{ab(a-b)^2+2(ab-1)}{(a^2+1)(b^2+1)(ab+1)}$
Ta thấy: $a^2+1;b^2+1>0∀a;b$; $a;b≥1⇒ab≥1⇒ab+1≥2>0$
$⇒(a^2+1)(b^2+1)(ab+1)>0(1)$
Mặt khác: $(a-b)^2≥0⇒ab(a-b)^2≥0;ab≥1$ và $ab≥1⇒ab-1≥0⇒2(ab-1)≥0$
$⇒ab(a-b)^2+2(ab-1)≥0∀a;b(2)$
Từ $1;2⇒\dfrac{ab(a-b)^2+2(ab-1)}{(a^2+1)(b^2+1)(ab+1)}≥0$
$⇒\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1} - \dfrac{2}{ab+1}≥0$
Hay $\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}≥\dfrac{2}{ab+1}$
Dấu $=$ xảy ra $⇔a-b=0;ab=1⇔a=b=1$
