A special inequality (related to 2021) Let $x,y,z\in[0,1]$ such that: $$x+y+z\leq 1+2xyz$$
I want to prove the following inequality (for a student in high school):
$$x^{2021}+y^{2021}+z^{2021}\leq 1+2\sqrt{xyz}^{2021}$$

Edit:
This question was sent to me by a student from high school, his teacher ask them to look for proof. I think it's a kind of question given in "International Mathematical Olympiad", that's why the "2021" number is used here. I tried by recurrence..
 A: The inequality is true. However, the only proof I know has nothing to do with high school nor IMO mathematics...
For $x,y,z \in [0,1]$, let $u = \sqrt{x}, v = \sqrt{y}, w = \sqrt{z}$. We have $1 - u^2, 1 - v^2, 1 - w^2 \ge 0$.
Together with
$$\left|\begin{matrix}1 & u & v\\ u & 1 & w\\ v & w & 1\end{matrix}\right| = 1-u^2-v^2-w^2+2uvw \ge 1 - x - y - z + 2xyz \ge 0$$
Sylvester's criterion tells us the matrix $M =\begin{bmatrix}1 & u & v\\ u & 1 & w\\ v & w & 1\end{bmatrix}$ is positive semi-definite.
For any integer $n > 1$, Schur product theorem tell us the $n$-fold Hadamard product of $M$ with itself is positive semi-definite. By Sylvester's criterion again, we find
$$1 - u^{2n} - v^{2n} - w^{2n} + 2(uvw)^n =
\left|\begin{matrix}1 & u^n & v^n\\ u^n & 1 & w^n\\ v^n & w^n & 1\end{matrix}\right| \ge 0$$
Substitute $n$ by $2021$, the desired inequality follows.
A: Alternative proof:
Let us prove that, for any integer $n \ge 2$,
$$x^n + y^n + z^n \le 1 + 2 \sqrt{xyz}^n.$$
Let
$$f(n) = 1 + 2\sqrt{xyz}^n - x^n - y^n - z^n.$$
We have
$$f(n + 1) - f(n) = 2\sqrt{xyz}^{n + 1} - x^{n + 1} - y^{n + 1} - z^{n + 1} - 2\sqrt{xyz}^n + x^n + y^n + z^n,$$
and
\begin{align*}
 &[f(n + 2) - f(n + 1)] - [f(n + 1) - f(n)]\\
 =\,\, & 2\sqrt{xyz}^n (1 - \sqrt{xyz})^2
 - x^n(1 - x)^2 - y^n(1 - y)^2 - z^n(1 - z)^2\\
 \le\,\,& 2\sqrt{xyz}^n (1 - \sqrt{xyz})^2 - 
 3\sqrt[3]{x^n(1 - x)^2 \cdot y^n(1 - y)^2 \cdot z^n(1 - z)^2}\\
 =\,\,& (xyz)^{n/3}
 \left[2(xyz)^{n/6} (1 - \sqrt{xyz})^2 - 3\sqrt[3]{(1 - x)^2(1 - y)^2(1 - z)^2}\right]\\
 \le\,\,& (xyz)^{n/3}
 \left[2(xyz)^{2/6} (1 - \sqrt{xyz})^2 - 3\sqrt[3]{(1 - x)^2(1 - y)^2(1 - z)^2}\right]\\
 \le\,\,& (xyz)^{n/3}
 \left[2(xyz)^{2/6} (1 - \sqrt{xyz})^2 - 2\sqrt[3]{(1 - x)^2(1 - y)^2(1 - z)^2}\right]\\
 \le\,\,& 0. \tag{1}
\end{align*}
(The proof of (1) is given at the end.)
Thus, we have
$$f(n + 2) - f(n + 1) \le f(n + 1) - f(n), \quad \forall n \ge 2.$$
Also, clearly $\lim_{n\to \infty} [f(n + 1) - f(n)]  = 0$.
Thus, we have
$$f(n + 1) - f(n) \ge 0, \quad \forall n\ge 2.$$
Also, we have
$f(2) = 1 + 2xyz - x^2 - y^2 - z^2 \ge 1 + 2xyz - x - y - z \ge 0$.
Thus,
$$f(n) \ge f(2) \ge 0, \quad \forall n \ge 2.$$
We are done.


Proof of (1):
It suffices to prove that
$$\sqrt{xyz} (1 - \sqrt{xyz})^3 \le (1 - x)(1 - y)(1 - z).$$
Let $p = x + y + z, q = xy + yz + zx, r = xyz$.
We have $1 + 2r \ge p$.
Using $q^2 \ge 3pr$, we have
\begin{align*}
 &(1 - x)(1 - y)(1 - z)\\
 =\,\,& 1 - xyz + (xy + yz + zx) - (x + y + z)\\
 =\,\,& 1 - r + q - p\\
 \ge\,\,& 1 - r + \sqrt{3pr} - p\\
 =\,\,& 1 - r - \left(\sqrt{p} - \frac12\sqrt{3r}\right)^2 + \frac34 r\\
 \ge\,\,& 1 - r - \left(\sqrt{1 + 2r} - \frac12\sqrt{3r}\right)^2 + \frac34 r
\end{align*}
where we have used $1 + 2r \ge p$ and $\sqrt{p} \ge \frac{1}{2}\sqrt{3r}$.
It suffices to prove that
$$\sqrt{r} (1 - \sqrt{r})^3 \le 1 - r - \left(\sqrt{1 + 2r} - \frac12\sqrt{3r}\right)^2 + \frac34 r$$
or
$$\sqrt{r}\left(\sqrt{3 + 6r} - 1 - 3r + r\sqrt{r}\right)\ge 0$$
which is true (using $r\in [0, 1]$).
We are done.
