bất đẳng thức khó quá

B

braga

$\text{Xét:} \ \dfrac{1}{1 + a^2} + \dfrac{1}{1 + b^2} - \dfrac{2}{ab + 1} \\ = \left(\dfrac{1}{1 + a^2} - \dfrac{1}{ab + 1}\right) + \left(\dfrac{1}{1 + b^2} - \dfrac{1}{ab + 1}\right) \\ = \dfrac{ab - a^2}{(1 + ab)(1 + a^2)} + \dfrac{b^2 - ab}{(1 + ab)(1 + b^2)} \\ = \dfrac{b - a}{1 + ab}.\left(\dfrac{a}{1 + a^2} - \dfrac{b}{1 + b^2}\right)
\\ = \dfrac{(b - a)(ab^2 - a^2b + a - b)}{(1 + a^2)(1 + b^2)(1 + ab)} \\ = \dfrac{(b - a)^2(ab - 1)}{(1 + a^2)(1 + b^2)(1 + ab)} \ge 0$
 
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