I have to prove or show that for $a \in\mathbb{R}$ if $x(x-(2a)^2)>0$, it follows that $|x-(a)^2|>a^2$ is true or not. I think it is true, as $x-(2a)^2$ has to be either bigger or as smaller than $0$, and it's also the case in the following situation but I'm not sure if there would be more cases when $x>0$ and $x<0$ is and therefore could be shown that it's false. I would be happy if you could tell me if my way is true or not.
I assumed that $x$ $\neq$ 0 and $x-(2a)^2$ $\neq$ 0. Then I calculated the inequality of $|x-(a)^2|>a^2$. $|x-(a)^2|>0$ and $|x-(a)^2|<0$. Then I showed that this preposition is true, as I assumed that $x-(2a)^2$ $\neq$ 0
Let $x(x-4a^2)>0$.
If $x>0$, $x - 4a^2>0$. Thus $x - 2a^2 > 2a^2 \ge 0$, this is, $x-a^2 > a^2 \ge 0$. Since $x - a^2 > 0$, $|x - a^2| = x - a^2$ and $|x - a^2| > a^2$.
If $x < 0$, $-x > 0$. Thus $a^2 - x > a^2 \ge 0$. Since $a^2 - x > 0$, $|a^2 - x| = a^2 - x$, but $|a^2 - x| = |x - a^2|$ and $|x - a^2| > a^2$.