How to prove that $\min(f(x)) = \max(1/f(x))$ Given a function $f(x)$ and the fact that $\frac{f(x)}{|f(x)|} = \text{const}$ (the sign of a fuction is constant) how can I prove that:
$$
\max(f(x)) = \min\left(\frac{1}{f(x)}\right)
$$
and the opposite:
$$
\max\left(\frac{1}{f(x)}\right) = \min(f(x))?
$$
My thoughts:
Given a function $f(x)$:
$$
\min(f(x)) \le f(x) \le \max(f(x)) \\
\min\left(\frac{1}{f(x)}\right) \le \frac{1}{f(x)} \le \max\left(\frac{1}{f(x)}\right)
$$
Raising $\min(f(x)) \le f(x)$ to a $-1$ power gives:
$$
\frac{1}{\min(f(x))} \ge \frac{1}{f(x)}.
$$
At this point it feels like $\frac{1}{\min(f(x))}$ is equal to $\max{\frac{1}{f(x)}}$. But how can this be formally proved? 
 A: The most important thing here is the initial statement that f(x)\|f(x)|=const. You can easily check that functions like $f(x)=x$ min and max will have different sign if the interval on which you evaluate it includes 0(or other points where function changes sign).
Which makes statement false at the start.
So, we know that f is or strictly positive or strictly negative always. So when you raise to -1 degree you actually raise two positive or negative numbers obtaining the inequality you provided.
Because imagine $$-5\leq 1$$ but then to -1 degree you obtain $$-1/5\leq 1$$ and not the opposite. That is why the sign equality is crucial here.
Another important fact is for proving it you need to be sure that maximum and minimum actually exist.
So if you have obtained the fact that minimum and maximum really exist and they are same sign you can continue with it using common sense:
Consider sign of the function +(for - its all the same) 
Then let $x_0$ be the point of maximum and so $max(f(x))=f(x_0)$. Then lets find minimum of (1/f(x)). Obviously as 1/x on positive arguments is strictly decreasing then minimum will be at the point of maximum of f(x).
So, the minimum will be obtained at the point $x_0$. But it is not equal to max as max is actual value and you were most probably looking for argmax. Argmax is basically an argument which provides function with the maximum value.(you can find info about it here https://en.wikipedia.org/wiki/Arg_max)
After this improvements we can finally say that $$argmax(f(x))=argmin(1/f(x))$$
