Áp dụng bất đẳng thức $Cauchy- Schwarz$ dạng $Engel$ ta được:
$\dfrac{x^4}{a} + \dfrac{y^4}{b} \geq \dfrac{(x^2 + y^2)^2}{a + b} = \dfrac{1}{a + b}$
Dấu = xảy ra $\Leftrightarrow \begin{cases}\dfrac{x^2}{a} = \dfrac{y^2}{b}\\x^2 + y^2 = 1\end{cases}$
$\Leftrightarrow \begin{cases}x^2 = \dfrac{a}{a + b}\\y^2 = \dfrac{b}{a + b}\end{cases}$
Do đó:
$\dfrac{x^{2020}}{a^{1010}} + \dfrac{y^{2020}}{b^{1010}}$
$= \left(\dfrac{x^2}{a}\right)^{1010} + \left(\dfrac{y^2}{b}\right)^{1010}$
$= 2.\left(\dfrac{x^2}{a}\right)^{1010}$
$= 2.\left(\dfrac{\dfrac{a}{a + b}}{a}\right)^{1010}$
$= 2.\left(\dfrac{1}{a + b}\right)^{1010}$
$= \dfrac{2}{(a + b)^{1010}}$
