c1:đặt $\frac{x}{y+1}=a $
$\frac{y}{x+1}=b$
\begin{cases}a^2+b^2=\frac{1}{2} \\ 4ab=1 \end{cases}
hệ này giải bt
c2:(1)$\frac{x^2}{(y+1)^2}+\frac{y^2}{(x+1)^2}\geq 2\left| {\frac{xy}{(x+1)(y+1)}} \right|$
(2) $\frac{xy}{(x+1)(y+1)}=1/4$
=> $(1)\geq 1/2$=> $\frac{x}{y+1}=\frac{y}{x+1}$