$bdt\Leftrightarrow \frac{2a}{a^2+1}+\frac{2b}{b^2+1}=\frac 12+\frac{2}{c^2+1}$
Đổi biến $(a,b,c) \longrightarrow \left( \tan \frac{A}2 ,\tan \frac B2, \tan \frac C2\right) \; \; (A+B+C = \pi)$
$bdt\Leftrightarrow 2\cos^2 \frac A2.\tan \frac A2+2\cos^2 \frac B2.\tan \frac B2 \le \frac 12+2\cos^2\frac C2$
$\Leftrightarrow \sin A+\sin B \le 2 \cos^2 \frac C2+\frac 12$
BDT cuối đúng do $\sin A+\sin B=2.\sin \frac{A-B}2.\cos \frac{A+B}{2} \le 2.1.\cos\frac{ (\pi-A-B)}{2}=2\cos \frac C2 \le VP$
Đẳng thức xảy ra $\Leftrightarrow A=B= \frac{\pi}6,C=\frac{2\pi}{3}$
hay $a=b=2-\sqrt 3,c=\sqrt 3$