<s>[INST] 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 $x_{max}=3.$
 [/INST] 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 $x\rightarrow+\infty.$
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