`a-\sqrta=\sqrtb-b`
`⇔a+b=\sqrta+\sqrtb\ge (\sqrta+\sqrtb)^2/2`
`⇔a+b=\sqrta+\sqrtb\le 2`
`P=a^2+b^2+2020/(\sqrta+\sqrtb)^2`
`P= (a+b)^2+2020/(a+b)^2-2ab`
`P=505/4[(a+b)^2+16/(a+b)^2]-2ab-[501(a+b)^2]/4`
`P\ge 505/4. 2\sqrt[(a+b)^2. 16/(a+b)^2]-2.1-501.4/4`
`P\ge 1010-2-501=507`
Dấu `=` xảy ra $\begin{cases}a=b\\a+b=\sqrt{a}+\sqrt{b}=2\\(a+b)^2=\dfrac{16}{(a+b)^2}\end{cases}⇔a=b=1$
Vậy $Min_P=507⇔a=b=1$