1) $n^5 - 2020n^3 + 4n$

$= n^5 - n - 2020n^3 + 5n$

$= n(n^4- 1) - 5n(404n^2 - 1)$

$= n(n^2-1)(n^2 +1) - 5n(404n^2 - 1)$

$= n(n-1)(n+1)(n^2 - 4 + 5) - 5n(404n^2 - 1)$

$= n(n-1)(n+1)(n^2-4) + 5n(n-1)(n+1) - 5n(404n^2 - 1)$

$= (n-2)(n-1)n(n+1)(n+2) + 5n(n-1)(n+1) - 5n(404n^2 - 1)$

Ta có:

$(n-2)(n-1)n(n+1)(n+2)$ là tích của `5` số nguyên liên tiếp

nên $(n-2)(n-1)n(n+1)(n+2)\quad \vdots \quad 5\qquad (1)$

Dễ dàng nhận thấy:

$\quad \begin{cases}5n(n-1)(n+1) \quad \vdots \quad 5\\5n(404n^2 - 1) \quad \vdots \quad 5\end{cases}$

$\Rightarrow 5n(n-1)(n+1) - 5n(404n^2 - 1) \quad \vdots \quad 5\qquad (2)$

$(1)(2) \Rightarrow (n-2)(n-1)n(n+1)(n+2) + 5n(n-1)(n+1) - 5n(404n^2 - 1) \quad \vdots \quad 5$

hay $n^5 - 2020n^3 + 4n \quad \vdots \quad 5\qquad (đpcm)$

2a) Ta có:

$x^2 + y^2 + z^2 \geq xy + yz + zx$

$\Leftrightarrow 2x^2 + 2y^2 + 2z^2 \geq 2xy + 2yz + 2zx$

$\Leftrightarrow 2x^2 + 2y^2 + 2z^2-2xy - 2yz - 2zx \geq 0$

$\Leftrightarrow (x^2 - 2xy + y^2) + (y^2 - 2yz + z^2) + (z^2 - 2zx + x^2) \geq 0$

$\Leftrightarrow (x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0$ (luôn đúng $\forall x,\,y,\,z$)

Vậy $x^2 + y^2 + z^2 \geq xy + yz + zx$

b) Áp dụng bất đẳng thức $Cauchy - Schwarz$ dạng $Engel$ ta được:

$x^2 + y^2 + z^2 = \dfrac{x^2}{1} + \dfrac{y^2}{1} + \dfrac{z^2}{1} \geq \dfrac{(x+y+z)^2}{1+1+1}$

$\Leftrightarrow x^2 + y^2 + z^2 \geq \dfrac{3^2}{3} = 3$

Dấu `=` xảy ra $\Leftrightarrow \begin{cases}x = y = z\\x + y + z = 3\end{cases}\Leftrightarrow  x = y = z = 1$

Vậy $\min(x^2 + y^2 + z^2) = 3 \Leftrightarrow x = y = z =1$