3. Solve the inequality $2021 \cdot \sqrt[202]{x^{2020}}-1 \geq 2020 x$ for $x \geq 0$. (10 points)
Solution. Transform the inequality into the form:

$$
\begin{aligned}
& \frac{2020 x+1}{2021} \leq \sqrt[202]{x^{2020}}, \text { from which } \\
& x+x+\ldots+x+1
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_66ff5449eae3e24affe1g-18.jpg?height=92&width=497&top_left_y=682&top_left_x=414)

But by the relation for the arithmetic mean and the geometric mean, $x+x+\ldots+x+1$

![](https://cdn.mathpix.com/cropped/2024_05_06_66ff5449eae3e24affe1g-18.jpg?height=100&width=1499&top_left_y=841&top_left_x=421)
the solution to the inequality is $x=1$.

Answer. $x=1$.