A special inequality (related to 2021)

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.

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.