Solve the following system of equations for $x, y \in \mathbb R$ - $\left\{ \begin{align} \sqrt x + \sqrt {y + 3} = 3\\ \sqrt{5 - x^2} + \sqrt y = 3 \end{align} \right.$.
$(x \in [0, \sqrt 5], y \ge 0)$
Of course, we could go further and have that $$\sqrt x + \sqrt {y + 3} = \sqrt{5 - x^2} + \sqrt y \iff \sqrt{5 - x^2} - \sqrt x = \sqrt{y + 3} - \sqrt y \le \sqrt 3$$
$$\implies x \in \left[-\frac{1}{2} + \sqrt 3 - \frac{1}{2}\sqrt{4\sqrt 3 - 3}, \sqrt 5\right]$$
Notwithstanding, it wouldn't help much.
Furthermore, we have that $$[(\sqrt{y + 3})^2 - (\sqrt y)^2] - (\sqrt{y + 3} + \sqrt y)(\sqrt{y + 3} - \sqrt y) = 0$$
$$\implies (3 - \sqrt x)^2 - (3 - \sqrt{5 - x^2})^2 - (6 - \sqrt x - \sqrt{5 - x^2})(\sqrt{5 - x^2} - x) = 0$$
$$\iff (\sqrt x - 1)\sqrt x(6 - \sqrt x - \sqrt{5 - x^2}) = 0 \iff \left[ \begin{align} \sqrt x - 1 = 0\\ \sqrt x = 0\\ 6 - \sqrt x - \sqrt{5 - x^2} = 0 \end{align} \right.$$
It needs to be proven that $6 - \sqrt x - \sqrt{5 - x^2} > 0$ for $\forall x \in [0, \sqrt 5]$, which I couldn't have in the three hour I was given in the competition.
$\implies \left[ \begin{align} x = 1\\ x = 0\\ \end{align} \right.$.
For $x = 1$, we have $y = 1$, and for $x = 0$, we have $y = 6$ and $y = \dfrac{\sqrt{10} - \sqrt 2}{2}$, which is, of course, illogical.
The only solution to the above system of equations is $x = y = 1$.
Is the above solution correct? How could I prove that $6 - \sqrt x - \sqrt{5 - x^2} > 0$ for $\forall x \in [0, \sqrt 5]$? And should you come up with any better solutions, please write them down below. Thanks for your attention.
No, it isn't.
Note that $\sqrt{y+3}-\sqrt y$ is equal to $\sqrt{5-x^2}-\color{red}{\sqrt x}$, not $\sqrt{5-x^2}-x$.
More importantly, $$(3 - \sqrt x)^2 - (3 - \sqrt{5 - x^2})^2 - (6 - \sqrt x - \sqrt{5 - x^2})(\sqrt{5 - x^2} - \color{red}{\sqrt x}) = 0$$ holds for every $x$ such that $x\in[0,\sqrt 5]$. (You can see by expanding LHS that LHS is a constant function which is equal to $0$.)
So, your method does not work.
Eliminating $y$, we get
$$(3-\sqrt x)^2-(3-\sqrt{5-x^2})^2=3$$ which is equivalent to $$ -(5-x^2)+6\sqrt{5-x^2}-8+(x-6\sqrt x+5)=0$$ which can be written as $$ (-\sqrt{5-x^2}+2)(\sqrt{5-x^2}-4)+(\sqrt x-1)(\sqrt x-5)=0$$ which is equivalent to
$$\frac{(x-1)(x+1)}{2+\sqrt{5-x^2}}(\sqrt{5-x^2}-4)+\frac{x-1}{\sqrt x+1}(\sqrt x-5)=0,$$ i.e.
$$(x-1)\bigg(\underbrace{\frac{x+1}{2+\sqrt{5-x^2}}}_{\text{positive}}(\underbrace{\sqrt{5-x^2}-4}_{\text{negative}})+\underbrace{\frac{1}{\sqrt x+1}}_{\text{positive}}(\underbrace{\sqrt x-5}_{\text{negative}})\bigg)=0$$
from which we get $x=1$ and $y=1$.