Find maximize and minimize of $P=x+y$

Question

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 $P=x+y$

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I found the maximize but minimize I have no idea. Help me.

Answer 1

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!

Answer 2

HINT:

Clearly, $y+1,x+1\ge0$

Let $\sqrt{x+1}=r\cos t,\sqrt{y+1}=r\sin t\implies x+y=r^2-2$

So, we have

$$r^2-r(\cos t+\sin t)-2=0$$

Now use $$-\sqrt2\le\cos t+\sin t\le\sqrt2$$