Help complete proof ('or' statement in conclusion): $a^2 \ge 7a \Rightarrow a\le0 \text { or } a\ge7$

Question

Question

Prove that if $a$ is a real number such that $a^2 \ge 7a$ then $a\le0 \text { or } a\ge7$

My Attempt

We are given:

1. $a \in \mathbb{R}$
2. $a^2 \ge 7a$

And need to prove:

$a\le 0 \text{ or } a\ge 7$ is true

Since:

$a\le 0 \text{ or } a\ge 7$

is logically equivalent to:

$a\lt 7 \Rightarrow a\le 0$

We can say that we are given:

1. $a\in \mathbb{R}$
2. $a^2 \ge 7a$
3. $a\lt 7$

And we need to prove:

$a\le 0$ is true

I am stuck here. I don't know what else to do complete this proof.

On the other hand, I also know that that "$a\le 0 \text{ or } a\ge 7$" is also logically equivalent to "$a>0 \Rightarrow a\gt 7$" and from here its clear to me how to complete the proof.

I'd like to know how to complete the proof from the point I am stuck at and the mental process that one goes through, as the problem I have with these kind of questions is that although I know the structure of the proof techniques, it is often the case that I am not able to immediately figure out how to arrive at the conclusion from the given statements. So, please also advise what can I do to fix this.

Answer 1

$a^2\ge7a\leftrightarrow a^2-7a\ge0\leftrightarrow a(a-7)\ge0$

The multiplication of two reals can only and only if non-negative, if

• both of them is non-negative (thus $a\ge0$ and $a-7\ge0$ $\leftrightarrow \underline{\underline{a\ge7}}$)
• or both of them is non-positive (thus $a\le0$ and $a-7\le0$ $\leftrightarrow \underline{\underline{a\le0}}$)

Answer 2

Perhaps an easier way is to notice the following: $$a^2 \geq 7a \iff a^2 - 7a \geq 0 \iff a(a-7) \geq 0$$ Now we only need to use the fact that $ab \geq 0$ iff $a, b \leq 0$ or $a, b \geq 0$.

Answer 3

the negative case is pretty obvious, the positive case: $a^2\geq7a \iff a\geq7$