<s>[INST] Math contest question - prove unsolvability of equations. Prove that the system of equations has no real solutions:
$$\begin{cases} y=\sqrt{x+\sqrt{1-x}} \\ x=\sqrt{y-\sqrt{1+y}}\end{cases}. $$
This is a former problem from a national math contest which I´ve solved already. However, since my solution was very similar to the solution manual, I´m interested in other solving approaches. 
My idea, in short, was as follows:
We first notice that from equation (1), we get the condition $y < \sqrt{2}$. 
From equation (2) we have that $y \geq \sqrt{1+y}$, but for $y<1$ this can't be true since then $\sqrt{1+y}>1$. So $y \geq 1$ and thus $$0 \leq y-\sqrt{1+y}\leq y-\sqrt{2},$$
which is a contradiction to our first condition $y < \sqrt{2}$.
 [/INST] Suppose that there exists a set of real solution.
First of all, $x$ has to satisfy the followings : 
$$1-x\ge 0\ \ \ \text{and}\ \ \ x+\sqrt{1-x}\ge 0\ \ \ \text{and}\ \ \ x\ge 0.$$
This is equivalent to $0\le x\le 1$.
Similarly, $y$ has to satisfy the followings : 
$$1+y\ge 0\ \ \ \text{and}\ \ \ y-\sqrt{1+y}\ge 0\ \ \ \text{and}\ \ \ y\ge 0.$$
This is equivalent to $y\ge\frac{1+\sqrt 5}{2}$.
Now let $f(x)=\sqrt{x+\sqrt{1-x}}\ \ \ (0\le x\le 1)$. Since we have
$$f'(x)=\frac{2\sqrt{1-x}-1}{4\sqrt{1-x}\cdot\sqrt{x+\sqrt{1-x}}},$$
we have $f'(x)=0\iff x=\frac 34$, and so $1=f(0)=f(1)\le f(x)\le f(3/4)=\frac{\sqrt 5}{2}$.
However, this contradicts $y\ge\frac{1+\sqrt 5}{2}$. 
</s>