$\dfrac{2x}{x^2+1}+\dfrac{2y}{y^2+1}+\dfrac{z^2-1}{z^2+1}\le \dfrac{3}{2}$

Bài toán:
Cho $x,y,z$ thực không âm thỏa mãn $xy+yz+zx=1$
Chứng minh rằng : 
$$\dfrac{2x}{x^2+1}+\dfrac{2y}{y^2+1}+\dfrac{z^2-1}{z^2+1}\le \dfrac{3}{2}.$$



Lời giải:
Cách 1:
Có:
$$(1+x^2)(1+y^2)=(xy-1)^2+(x+y)^2=(yz+zx)^2+(x+y)^2$$
$$=(z^2+1)(x+y)^2.$$
Bđt trở thành:
$$\begin{align*} &\dfrac{2[x(y^2+1)+y(x^2+1)]}{(x^2+1)(y^2+1)}-\dfrac{2(x+y)^2}{(z^2+1)(x+y)^2}\le \dfrac{1}{2}\\ \Leftrightarrow &4(xy+1)(x+y)-4(x+y)^2\le (x^2+1)(y^2+1)\\ \Leftrightarrow &(x-y)^2+(xy+1-2x-2y)^2\ge 0.~\text{(LĐ)}. \end{align*}$$
Dấu "=" xảy ra khi $x=y=2-\sqrt{3};z=\sqrt{3}$.

Cách 2:

Vì $xy+yz+zx=1$ nên ta có thể đặt $x=\tan\dfrac{A}{2};y=\tan\dfrac{B}{2};z=\tan\dfrac{C}{2}$

$$\Leftrightarrow \sin A+\sin B-\cos C\le \dfrac{3}{2}$$
$$\Leftrightarrow -2\cos^2 \dfrac{C}{2}+2\cos \dfrac{A-B}{2}\cos \dfrac{C}{2}-\dfrac{1}{2}\le 0.$$

Ta có 

$$\Delta '=\cos^2\dfrac{A-B}{2}-1\le 0.$$

nên BĐT luôn đúng.

Dấu "=" xảy ra khi $\left\{\begin{matrix} \cos\dfrac{A-B}{2}=1 & \\ \cos\dfrac{C}{2}=\dfrac{1}{2} & \end{matrix}\right.\Leftrightarrow A=B=\dfrac{\pi}{6};C=\dfrac{2\pi}{3}\Leftrightarrow x=y=2-\sqrt{3};z=\sqrt{3}.$