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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?

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    @ShrevatsaR: Agreed, much better :)2019-01-24

5 Answers 5

13

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.

12

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 $a^2 < b^2$.

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    Glad you found this answer useful.2011-07-21
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3

If $0 then both $b-a>0$, $b+a>0$. This implies that

$b^2-a^2=(b-a)(b+a)>0,$

which is equivalent to $b^2>a^2$.

2

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).