Taking derivatives, we wish to find when $$\frac{\sqrt{2020-x}-\sqrt{x-2018}}{2\sqrt{(x-2018)(2020-x)}}$$is equal to zero or undefined. This is equal to zero iff $x=2019$, and is undefined iff $x=2018,2020$. Plugging these values in, we see that the expression is equal to $\sqrt{2}$ is $x=2020$ or $2018$, and is equal to $2$ when $x=2019$. Thus, the maximum is $\boxed{2}$, achieved when $x=2019$.

Solution

Wouldn’t $\frac{d}{dx} \sqrt{x-2018}+\sqrt{2020-x}$ $=\frac{1}{2\sqrt{x-2018}}-\frac{1}{2\sqrt{2020-x}}$. Setting this equal to zero, we get $x-2018=2020-x$, so $x=2019$

what is QM-AM? oh RMS-AM inequality

https://artofproblemsolving.com/wiki/index.php/Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality

Directly apply cauchy shwarz inequality and the answer is easy to find.

Using the Cauchy-Schwarz inequality, we get that: $(\sqrt{x-2018}^2 + (\sqrt{2020-x}^2)(1^2 + 1^2) \geq (\sqrt{x-2018} + \sqrt{2020-x})^2$ $(x-2018+2020-x) \cdot 2 \geq (\sqrt{x-2018} + \sqrt{2020-x})^2$ $2^2 \geq (\sqrt{x-2018} + \sqrt{2020-x})^2 \Rightarrow 2 \geq \sqrt{x-2018} + \sqrt{2020-x}$ Therefore, the maximum value is $2$ $\blacksquare$

Let $M=\sqrt{x-2018} + \sqrt{2020-x}$ $\implies$ $M^2=(\sqrt{x-2018} + \sqrt{2020-x})^2=x-2018+2\sqrt{(x-2018)(2020-x)} +2020-x=x-x+2020-2018+2\sqrt{(x-2018)(2020-x)}$ $=2+2\sqrt{(x-2018)(2020-x)}$ $\implies$ $M^2=2+2\sqrt{(x-2018)(2020-x)}$ By $AM-GM$ we get: $M^2=2+2\sqrt{(x-2018)(2020-x)} \leq 2+(x-2018)+(2020-x)=2+x-2018+2020-x=2+x-x+2020-2018=2+2=4$ $\implies$ $M^2 \leq 4$ $\implies$ $M \leq 2$ $\implies$ $\sqrt{x-2018} + \sqrt{2020-x} \leq 2$ So the maximum of $\sqrt{x-2018} + \sqrt{2020-x}$ is $2$ And the equality holds when $x-2018=2020-x$ $\implies$ $2x=4038$ $x=2019$ So the maximum of $\sqrt{x-2018} + \sqrt{2020-x}$ is $2$ for $x=2019$