Proof by cases, inequality I have the following exercise:

For all real numbers $x$, if $x^2 - 5x + 4 \ge 0$, then either $x \leq 1$ or $x \geq 4$. 

I need you to help me to identify the cases and explain to me how to resolve that. Don't resolve it for me please.

You already have the cases so I'm confused about the question. I assume you are asking how one gets to this. So let's focus on $x^2-5x+4$. We can factor this as

$$x^2-5x+4=(x-1)(x-4)$$

So what we really have is...

 $$(x-1)(x-4)\geq 0$$

Notice that if $x\geq 4$, both $x-1$ and $x-4$ are not negative. So their product will be positive (or $0$). Now if $x \leq 1$, we know that $x-1$ and $x-4$ are both negative (or 0) so their product is either $0$ or a positive number. I'll leave it to you to check that if $1 < x < 4$ that their product is negative and hence doesn't satisfy what you want.
