<s>[INST] Find maximize and minimize of $P=x+y$ 
For $\{x,y\}\subset\mathbb R$ such that $\sqrt{x+1}+\sqrt{y+1}=\sqrt{2}\left(x+y\right)$ find maximize and minimize of $P=x+y$


I found the maximize but minimize I have no idea. Help me. 
 [/INST] By AM-GM
$$2(x+y)^2=x+y+2+2\sqrt{(x+1)(y+1)}\leq x+y+2+(x+1+y+1)=2(x+y+2).$$
Thus, $$(x+y)^2-(x+y)-2\leq0$$ or
$$-1\leq x+y\leq2.$$
In the right inequality the equality occurs for $x=y=1$, 
which says that $2$ is a maximal value of $P$.
Now, $\sqrt{.}$  is a concave function.
Thus, since $(x+1+y+1,0)\succ(x+1,y+1)$, by Karamata we obtain:
$$\sqrt2(x+y)=\sqrt{x+1}+\sqrt{y+1}\geq\sqrt{0}+\sqrt{x+1+y+1}$$ or
$$2(x+y)^2-(x+y)-2\geq0$$ or
$$x+y\geq\frac{1+\sqrt{17}}{4}.$$
The equality occurs for $y=-1$ and $x=\frac{5+\sqrt{17}}{4}$, 
which says that $\frac{1+\sqrt{17}}{4}$ is a minimal value.
Done!
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