Proof that maximizing a function is equivalent to minimizing its negative

$f$ is maximized at $x$ if $f(x)\geq f(y)$, for all $y$. Similarly, $−f$ is minimized at $x$ if $−f(x) \leq −f(y)$, for all $y$.

Let's now add to the left and right sides of the inequality $−f(x) \leq −f(y)$ respectively $f(y)-f(y) = 0$ and $f(x)-f(x) = 0$, that is

$$(f(y)-f(y))-f(x) \leq (f(x)-f(x))-f(y)$$

which can be written as

$$f(y) + (-f(y)-f(x)) \leq f(x) + (-f(x)-f(y))$$

we can remove $(-f(y)-f(x))$ from both sides to obtain

$$f(y) \leq f(x) $$

This proof followed from the ordering property of real numbers.

HINT: use the definition of global maximum
Given $f:X \rightarrow R$, $x_0$ is a global maximum if $\forall x \in X, f(x)\leq f(x_0)$