From baby Rudin (exercise 5 in chapter 1):
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x\in A$. Prove that $$\inf A=-\sup (-A).$$
I tried to prove this in the following way, but I'm not sure if it is an actual valid proof:
Let $\alpha=\sup(-A)$ (it exists since $A$ being bounded below implies $-A$ being bounded above. This means that for all $x\in A$, $\alpha\geqslant-x$ holds (since $-x\in(-A)$). Multiplying both sides by $-1$, we get $-\alpha\leqslant x$. This proves that $-\sup (-A)=\alpha$ is a lower bound of $A$.
Now let's say we have $\beta>\alpha$. Yet again multiplying both sides by $-1$, we aquire $-\beta<-\alpha$. Since $-\alpha=\sup(-A)$, there must exist $x\in A$ such that $-\beta<x$. Multiplying both sides by $-a$, we get $\beta>-x$. But $-x\in(-A)$, so $\beta$ is not a lower bound of $-A$. Therefore, $\alpha$ must be the greatest lower bound of $A$.