Inequality for grade 10

Here's my Vieta approach, rewrite the condition $x+y=\sqrt{x+6}+\sqrt{y+6}$ Now we set $a=\sqrt{x+6}\geq 0, b=\sqrt{y+6}\geq 0$ , then $\begin{cases} a^2+b^2-12=a+b \\ a^2+b^2-12=P\end{cases}$ $\Leftrightarrow\begin{cases} a+b=P \\ ab=\frac{1}{2}(P^2-P-12)\end{cases}$ So $a,b$ will be the roots of $X^2-PX+\frac{1}{2}(P^2-P-12)=0$ In order for the above equation to have 2 non-negative roots $\begin{cases}\Delta\geq 0\\ a+b\geq 0\\ ab\geq 0\end{cases}$ $\Leftrightarrow\begin{cases} P^2-2P-24\leq 0 \\ P^2-P-12\geq 0\\ P\geq 0\end{cases}$ $\Rightarrow\begin{cases} P\in[-4;6] \\ P\in (-\infty;-3]\cup [4;+\infty) \\ P\geq 0\end{cases}$ So we have $4\leq P\leq 6$ and the product is $24$ , to find the equality cases, we subtitute the value of $P$ into the equation and solve for $x,y$

$P=x+y=\sqrt{x+6}+\sqrt{y+6}$ so that $P=x+y> 0$

For the maximum value,by applying C-S inequality we have:

$P^{2}=(x+y)^2=(\sqrt{x+6}+\sqrt{y+6})^2\leq 2(x+y+12)=2P+24$ And $P^2-2P-24\leq0 \Rightarrow (P-6)(P+4)\leq 0 \Leftrightarrow P\leq 6$

The equality holds when $x=y=3$

For the minimum value,by applying this inequality: $\sqrt{A}+\sqrt{B}\geq \sqrt{A+B}$

$P=x+y=\sqrt{x+6}+\sqrt{y+6}\geq \sqrt{x+y+12}\Rightarrow P\geq \sqrt{P+12}$

And $P^2-P-12\geq 0\Leftrightarrow (P-4)(P+3)\geq 0\Rightarrow P\geq 4$

The equality holds when $x=10;y=-6$ or it's permutation.

So that $4\leq P\leq 6$ and it's product is $24$