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.
 A: Hint: If $x^2 -5x + 4 \geq 0$ then $(x-4)(x-1) \geq 0$.
Hint 2: By trichotomy, we know that either $x>0$, $x=0$, or $x<0$. Consider these three cases. Then It is clear that either $x>0$ or $x<0$. Now what happens if $x>0$ or $x<0$? How can you still satisfy the inequality? Try playing with some values...
A: HINT:
If $(x-a)(x-b)\ge0$  
Now the product of two terms is $\ge0$
So, either both $\ge0$ or both $\le0$
Now in either case, find the intersection of the ranges of $x$
A: 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.
