TITLE: Showing $a^2 < b^2$, if $0 < a < b$
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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).