Irrational Equations

Firstly, we know that $1+2xy>0$ for $\frac{2}{\sqrt{1+2xy}}$ to be real.

Since $\sqrt{x(1-2x)}, \sqrt{y(1-2y)} \geq 0$ , we have $x(1-2x), y(1-2y) \geq 0$ . By solving them, we obtain

$0 \leq x, y \leq \frac{1}{2}.$

We now proceed to show that for $0 \leq x, y \leq \frac{1}{2}$ ,

$\frac{1}{\sqrt{1+2x^2}}+\frac{1}{\sqrt{1+2y^2}} \leq \frac{2}{\sqrt{1+2xy}} \cdots (1)$

In fact,

$(1) \Leftrightarrow \frac{1}{1+2x^2}+\frac{1}{1+2y^2}+\frac{2}{\sqrt{(1+2x^2)(1+2y^2)}} \leq \frac{4}{1+2xy}$

$\Leftrightarrow \left(\frac{1}{1+2x^2}+\frac{1}{1+2y^2}-\frac{2}{1+2xy}\right)+\left(\frac{2}{\sqrt{4x^2y^2+2x^2+2y^2+1}}-\frac{2}{2xy+1}\right) \leq 0$

$\Leftrightarrow\left[-\frac{2x(x-y)}{(1+2x^2)(1+2y^2)}+\frac{2y(x-y)}{(1+2y^2)(1+2xy)}\right]+\left(\frac{2}{\sqrt{4x^2y^2+2x^2+2y^2+1}}-\frac{2}{2xy+1}\right) \leq 0$

$\Leftrightarrow -\frac{2(x-y)^2(1-2xy)}{(1+2xy)(1+2x^2)(1+2y^2)}+\left(\frac{2}{\sqrt{4x^2y^2+2x^2+2y^2+1}}-\frac{2}{2xy+1}\right) \leq 0$

On the left hand side, the first term is clearly less than or equal to $0$ , and for the second term,

$\frac{2}{\sqrt{4x^2y^2+2x^2+2y^2+1}} \leq \frac{2}{\sqrt{4x^2y^2+4xy+1}}=\frac{2}{2xy+1}$

This implies that it is also less than or equal to $0$ , so $(1)$ is proven, and equality holds if and only if $x=y$ .

Hence, the second equation becomes

$\sqrt{x(1-2x)}=\sqrt{y(1-2y)}=\frac{1}{9},$

which yields $x=y=\frac{9\pm\sqrt{73}}{36}$ . Thus $a=9, b=73, c=36$ , and the desired answer is $\boxed{118}$ . $\square$

I see that you fixed the original posting $x\sqrt { x(1-2x) } +\sqrt { y(1-2y) } =\frac { 2 }{ 9 }$ . Now, finally, this problem has an answer.

First of all, if $y=x$ , then the top equation is obviously true. Then, letting $y=x$ , the 2nd equation quickly becomes $162{ x }^{ 2 }-81x+1=0$ , which leads directly to the solution.