If $f(x)^2$ is convex and $f(x)>1$, does $\arg\min_x f(x)^2=\arg\min_x f(x)$?

Question

Not sure, maybe it's trivial... Thought about that during my shower this morning.

My intuition is as follows. Let $x'=\arg\min_x f(x)^2$. For all $h$, we should have $$f(x+h)^2\ge f(x)^2$$ Then, since $f(x)>1$, we have $f(x)^2\ge f(x)$. Therefore, we also have $$f(x+h)^2\ge f(x)^2 \ge f(x)$$ And now.... I'm stuck.

Answer 1

Let $D$ the set of definition of $f$ and $x'=\arg\min_x f(x)^2$, then

$f(x')^2 \le f(x)^2$ for all $x \in D$.

Since $f>1$ on $D$, we have $f>0$ on $D$, hence

$f(x') \le f(x)$ for all $x \in D$.