Find min and max of $A = \sqrt{x-2} + 2\sqrt{x+1} + 2019 -x$ 
Find minimum and maximum of $$A = \sqrt{x-2} +  2\sqrt{x+1} + 2019 -x$$

How to solve this problem without using derivatives?

Michael Rozenberg's edit: 
Because with a derivative it's not so easy:
$$A'(x)=\frac{1}{2\sqrt{x-2}}+\frac{1}{\sqrt{x+1}}-1=\frac{1}{2\sqrt{x-2}}-\frac{1}{2}+\frac{1}{\sqrt{x+1}}-\frac{1}{2}=$$
$$=\frac{1-\sqrt{x-2}}{2\sqrt{x-2}}+\frac{2-\sqrt{x+1}}{2\sqrt{x+2}}=$$
$$=(3-x)\left(\frac{1}{2(1+\sqrt{x-2})\sqrt{x-2}}+\frac{1}{2(2+\sqrt{x+1})\sqrt{x+2}}\right),$$
which gives $x_{max}=3.$
 A: Let $f(x) = \sqrt{x-2} +  2\sqrt{x+1} + 2019 -x$ then
$$
f'(x) = \frac{1}{2\sqrt{x - 2}} + \frac{1}{\sqrt{x+1}} - 1
$$
Equating this to zero for real $x$, we get $x = 3$. The second derivative is negative hence the global maximum is $f(3) = 2021.$
A: By C-S and AM-GM we obtain:
$$\sqrt{x-2}+2\sqrt{x+1}-x+2019=\sqrt{x-2}+4\sqrt{\frac{x+1}{4}}-x+2019\leq$$
$$\leq\sqrt{(1+4)\left(x-2+4\cdot\frac{x+1}{4}\right)}-x+2019=\sqrt{5(2x-1)}-x+2019\leq$$
$$\leq\frac{5+2x-1}{2}-x+2019=2021.$$
The equality occurs for $x=3,$ which says that we got a maximal value.
The minimum does not exist. Try $x\rightarrow+\infty.$
A: $$ \forall_{y\ge -1}\quad\sqrt{1+y}\ \le\ 1+\frac y2 $$
hence
$$ \forall_{x\ge 2}\quad\sqrt{x-2}\ =\ \sqrt{1+(x-3)}\ \le\ 1 +\frac{x-3}2 $$
and
$$ \forall_{x\ge -1}\quad\sqrt{x+1}\ =\ 2\cdot\sqrt{1+\frac{x-3}4}\ \le
           \ 2+\frac{x-3}4 $$
hence
$$ \forall_{x\ge 2}\quad\sqrt{x-2}\ +\ 2\cdot\sqrt{x+1}\ \le\ 2+x $$
i.e.
$$ \forall_{x\ge 2}\quad\sqrt{x-2}\ +\ 2\cdot\sqrt{x+1}\ -\ x\ \le\ 2 $$
where the equality holds only for $\ x=3.\ $ Thus A (see the OP Question) attains its maximum $A=2021$ at $x=3$.   Great!
