Inequality ~ 1

Let $f(x) = \sqrt{x^2-4x+4} = \sqrt{(x-2)^2}=|x-2|$ , where $|\cdot|$ denotes the absolute value function . This means $f(x) \ge 0$ and $f(x)$ has a minimum value of 0 when $x=2$ .

Now, let $g(x) = x^2(x-2)$ . Since $x^2 \ge 0$ for all $x$ , $g(x) < 0$ , only when $x-2 < 0$ . Therefore, $g(x) \le 0$ for $x \le 2$ . Since $f(x) \ge 0$ for all $x$ and $f(2) = 0$ , $\implies f(x) \ge g(x)$ for $x \le 2$ . When $x> 2$ , $f(x) = |x-2| = x- 2$ . Then $g(x) = x^2(x-2) = x^2 f(x)$ . Since $x^2 > 4$ for $x > 2$ , $\implies g(x) > f(x)$ for $x > 2$ .

Therefore, $\sqrt{x^2 - 4x+4} \ge x^2(x-2)$ for $\boxed{x \le 2}$ .

It can be clearly seen by graphing the two functions together.

using absolute value $\sqrt{x^2}=|x|$