, 2009

The condition for the system is: $0 \le x, y \le \frac{1}{2}$. (*)

Remark: Under (*), we have
$$\frac{1}{\sqrt{1+2x^2}} + \frac{1}{\sqrt{1+2y^2}} \le \frac{2}{\sqrt{1+2xy}}$$
The equality holds iff $x = y$.

Proof: According to the Cauchy-Schwarz inequality, we have
$$\left( \frac{1}{\sqrt{1+2x^2}} + \frac{1}{\sqrt{1+2y^2}} \right)^2 \le 2 \left( \frac{1}{1+2x^2} + \frac{1}{1+2y^2} \right) \quad (1.1)$$
The equality holds iff $\sqrt{1+2x^2} = \sqrt{1+2y^2} \Leftrightarrow x=y$ (as $x, y \ge 0$).

Next, we have
$$\frac{1}{1+2x^2} + \frac{1}{1+2y^2} - \frac{2}{1+2xy} = \frac{2(y-x)^2(2xy-1)}{(1+2xy)(1+2x^2)(1+2y^2)} \le 0$$
Consequently:
$$\frac{1}{1+2x^2} + \frac{1}{1+2y^2} \le \frac{2}{1+2xy} \quad (1.2)$$
The equality holds $\Leftrightarrow x = y$.

(1.1) and (1.2) imply the inequality stated.
The equality holds iff the equality holds in both (1.1) and (1.2) that is $x = y$.

From Remark it follows that the given system is equivalent to the following
$$\begin{cases} x = y \\ \sqrt{x(1-2x)} + \sqrt{y(1-2y)} = \frac{2}{9} \end{cases} \Leftrightarrow \begin{cases} x = y = \frac{9 - \sqrt{73}}{36} \\ x = y = \frac{9 + \sqrt{73}}{36} \end{cases}$$
Thus we have two above two solutions.