Prove that $0\le a^2 < b^2$ implies $a < b$ Q) $a$ and $b$ are positive real numbers and $a^2 < b^2$. Prove that $a < b$
My working so far: Assuming always that $a, b > 0$, the contrapositive of $$a^2 < b^2 \implies a < b$$ is 
$$a \ge b \implies a^2 \ge b^2\;.$$
Thanks
 A: HINT: Multiply $a\ge b$ by $a$ to get $a^2\ge ab$ and by $b$ to get $ab\ge b^2$.
A: A more direct approach.
If $a,b>0$, then $a+b>0$, and hence $\frac{1}{a+b}>0$.
If $a^2<b^2$ then $$0 < b^2-a^2 = (b-a)(b+a)$$
Multiply both sides by $\frac{1}{a+b}$ to get $$0< b-a$$ so $a<b$.
This shows that you only really need $a+b>0$ and $b^2>a^2$ to show $b>a$.
A: $b\le a $ $\implies$ $b^2\le ab\le a^2$
A: I solve it in this way
if $a < b$
by multiplying both side by $a$ .. we get $a^2 < a.b$
,, also by multiplying both side by $b$ .. we get $a.b < b^2$
now we have $a^2 < a.b < b^2$
So $a^2 < b^2$ 
A: This might not be considered a formal proof but it is convincing (because of tis geometrical nature) plus generalises the fact to $(a^n < b^n \Rightarrow a < b \hspace{.5cm} \forall n \in \mathbb{N})$
Think of "$a^n < b^n$" as "the n-volume of a n-cube of side $a$ is smaller than the n-volume of a n-cube of side $b$"
This, as we are talking of n-cubes, certainly means that the n-cube with less volume fits into the other (imagine them sharing one corner) and so the side of the smaller one fits into the side of the bigger one which means $a<b$.
A: I know you probably want a trivial proof. But you can see this way: what you want is basically
 proof that $\sqrt{x}$ is crescent . The derived is positive $\implies$ $\sqrt{x}$ crescent
