Showing $a^2 < b^2$, if $0 < a < b$ Lately, I've been stumbling with proofs of inequalities.
For example:

Given $0 < a < b$
  Show $a^2 < b^2$

The only thing I've been able to come up with so far:

$a^2 < b^2$
  $\sqrt{a^2} < \sqrt{b^2}$
  $a < b$

OR


$a < b$
  $a^2 < b^2$

However, neither of these solutions seem to be really "showing" that $a^2 < b^2$, assuming $0 < a < b$. I've tried some other things, but to no avail. Am I merely overthinking the problem when, in fact, these are actually acceptable solutions, or am I truly missing something here?
 A: $$0<a<b\Rightarrow 0\cdot a<a\cdot a<b\cdot a<b\cdot b\Rightarrow
a^{2}<b^{2}.$$
A: If $0<a<b$ then both $b-a>0$, $b+a>0$.
This implies that
$$b^2-a^2=(b-a)(b+a)>0,$$
which is equivalent to $b^2>a^2$.
A: Specialize  $\rm\ A,B = a,b\ $ in the $ $ Inequality Product Rule $ $ below.
Lemma $\rm\ \ \color{#0a0}{b>a,\:B>A}\ \Rightarrow\ b\:B>a\:A\ $ if $ $ at most one of $\rm\ a,b,A,B\:$ is $\:\le 0$ 
Proof $\rm\quad bB\!-\!aA = \color{#c00}b\,(\color{#0a0}{B\!-\!A})+(\color{#0a0}{b\!-\!a})\color{#c00}A > 0\ $ by wlog $\rm\ a \le 0\ \Rightarrow\ \color{#c00}{b,A} > 0\:$ 
Note $\ $ The proof is essentially the same as the well-known proof of the Congruence Product Rule, a rule which lies at the heart of many product rules (e.g. for derivatives - see said post).
A: You have 
\begin{align*}
(a-b) &< 0  \\ \Longrightarrow (a-b) \cdot (a+b) &<0  \qquad\qquad \Bigl[\small\text{Multiplying both sides by}\ (a+b) \ \text{doesn't change the sign.} \Bigr]\\ \Longrightarrow a^{2}-b^{2} &< 0
\end{align*}
Or consider the function $f:\mathbb{R}_{>0} \to \mathbb{R}_{>0}$ given by $f(x) =x^{2}$. Clearly this monotonic because $f'(x) = 2x > 0$ for all $x >0$. Hence your claim follows.
A: Recall that multiplication by a positive number preserves order: since $a > 0$, then from $a < b$ you get $aa < ab$, and since $b > 0$, then from $a < b$ you get $ab < bb$. Together, this gives $aa < ab < bb$, hence $a^2 < b^2$.
