Inequality Proof Question is Confusing Because it Contains extra Variables I am trying to help my daughter with her homework.  I am confused by the problem, but I will admit that the math is a bit over my head.
Here is the beginning of the problem:

Let a $\in \mathbb{R}$. Prove that $a^2 \le 1$ if and only if $-1 \le  a \le 1$

I can more or less follow that and could maybe make an attempt at the proof.  But it goes on to say:

In the proof you may use the following two facts that are true for any a,b,c $\in \mathbb{R}$.

*

*If a < b and c > 0 then ac < bc

*If a < b and c < 0 then ac > bc

This has me really confused because the original problem did not have b or c in them.  How is adding facts about b and c relevant if the original inequality only includes a?
 A: Just because given facts use a variety of variables does not preclude some (or all) of them from being equal. As a more specific example, in your given fact 1, note that b ≠ a (otherwise we'd have a < a)  but nothing prevents c = a or c = b.
To see how that helps, suppose in fact 1, we let a = c = $a$ and b = $1$. Then fact 1 would become

*

*If $\,a<1\,$ and $\,a>0\,$ then $\,a^2<a\,$.

The antecedent can be rewritten as $\,0<a<1\,$; the consequent $\,a^2<a\,$ can be combined with the condition $\,a<1\,$ to yield $\,a^2<1\,$. Thus fact 1 gives us

*

*If $\,0<a<1\,$ then $\,a^2<1\,$
which is one part of what you're trying to prove.
One more example, in case negatives seem more confusing. Suppose in fact 2, we let a = $-1$ and b = c = $a$; fact 2 would become


*If $\,-1 < a\,$ and $\,a<0\,$ then $\,-a>a^2\,$.

We can again combine the two inequalities in the antecedent. For the consequent, note that $\,a>-1\,$ implies $\,-a<1\,$; combining that with the consequent $\,a^2<-a\,$ yields $\,a^2<1\,$. So fact 2 gives us


*If $\,-1<a<0\,$ then $\,a^2<1\,$
which is a second part of what you're trying to prove.
I think you'll be able to figure out what substitutions are needed to show that
if $\,a<-1\,$ then $\,a^2>1\,$
and
if $\,a>1\,$ then $\,a^2>1\,$
Combining all of the above with the simple fact that if $\,a=-1\,$ or $\,a=1\,$ then $\,a^2=1\,$ gives you the desired proof.
A: Suppose $a^2 \le 1$. We want to show that $-1 \le a \le 1$.
This is equivalent to showing that both $a>1$ and $a<-1$ are false.
Suppose for sake of contradiction that $a>1$. Multiply both sides by $c=a >0$ to get $a^2 > a$ and since $a>1$ we have $a^2 > 1$ which cannot be true.
A similar story applies for the other side.
Now suppose $-1 \le a \le 1$.
If $a = 1$ then we see that $1^2 =1\le 1$, so we can assume that $a \neq 1$. Starting with $a <1$ and letting $b=a$ we get $a^2 < a$ and since $a < 1$ we get
$a^2 < 1$ and so $a^2 \le 1$.
A similar story applies for the other side.
A: Alternative approach:
You are trying to find all values of $a$ such that $a^2 \leq 1.$
There are $3$ possibilities to consider:

*

*$a < 0$


*$a = 0$


*$a > 0$
So, you examine each of the $3$ constraints above, one at a time.  For each constraint, you find all values of $a$ that meet the constraint and also satisfy $a^2 \leq 1.$
You are expected to use the following facts:

*

*$a < b, c > 0 \implies ac < bc.$


*$a < b, c < 0 \implies ac > bc.$

$\underline{\text{Case 1} ~: ~a < 0}$
Let $c = a \implies c < 0$.
Then, $a < -1 \implies $

*

*Since $a = c < -1,~$ you have that $~c(-1) > (-1)(-1) = 1.$


*$a^2 = ac > (-1)c > 1.$
Also, $a = -1 \implies a^2 = (-1)^2 = 1.$
Finally, $0 > a = c > -1 \implies $

*

*Since $a = c > -1,~$ you have that $~c(-1) < (-1)(-1) = 1.$


*$a^2 = ac < (-1)c < 1.$
So, you have established that when $a < 0$, the only satisfying values are $-1 \leq a < 0.$

$\underline{\text{Case 2} ~: ~a = 0}$
Immediate that in this case, $a^2 = 0^2 < 1.$

$\underline{\text{Case 3} ~: ~a > 0}$
Let $c = a \implies c > 0$.
Then, $a > 1 \implies $

*

*Since $a = c > 1,~$ you have that $~c(1) > (1)(1) = 1.$


*$a^2 = ac > (1)c > 1.$
Also, $a = 1 \implies a^2 = (1)^2 = 1.$
Finally, $0 < a = c < 1 \implies $

*

*Since $a = c < 1,~$ you have that $~c(1) < (1)(1) = 1.$


*$a^2 = ac < (1)c < 1.$
So, you have established that when $a > 0$, the only satisfying values are $0 < a \leq 1.$

$\underline{\text{Putting This All Together}}$
The set of all satisfying values of $a$ are represented as the union of the following non-intersecting subsets:

*

*$a < 0$ and $-1 \leq a.$


*$a = 0$.


*$a > 0$ and $1 \geq a.$
Combining these three subsets produces:
$$a^2 \leq 1 \iff  [ ~(-1 \leq a < 0) ~~~~\text{or}~~~~ (0 = a) ~~~~\text{or}~~~~ (0 < a \leq 1) ~].$$
