For $x,y$. Minimize and Maximize $\left(x+y\right)^2-\sqrt{9-x-y}+\frac{1}{\sqrt{x+y}}$

Question

For $x,y$ satisfy $x+y-1=\sqrt{2x-4}+\sqrt{y+1}$, minimize and maximize $$\left(x+y\right)^2-\sqrt{9-x-y}+\frac{1}{\sqrt{x+y}}$$

Answer 1

$$S=\left(x+y\right)^2-\sqrt{9-x-y}+\frac{1}{\sqrt{x+y}}$$

We have $$\sqrt{2x-4}+\frac{1}{\sqrt{2}}\sqrt{2(y+1)}\leq\sqrt{3(x+y-1)}$$

From $$x+y-1=\sqrt{2x-4}+\sqrt{y+1}\Rightarrow x+y-1\leq\sqrt{3(x+y-1)}$$ Hence $$1\leq(x+y)\leq4$$

Let $$t=x+y;1\leq t\leq4$$ we have

$$S=t^2-\sqrt{9-t}+\frac{1}{\sqrt{t}}$$

Survey on function $S(t)$ on domain $[1;4]$ (covariates function)

We find $P_{max}=\frac{33-2\sqrt{5}}{2}$ when $x=4;y=0$

$P_{min}=2-2\sqrt{2}$ when $x=2;y=-1$