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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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    $\begingroup$ Yes, your first "proof" (it's actually backwards) uses the property that $a^2 < b^2 \implies \sqrt{a^2} < \sqrt{b^2}$, and the second one uses the property that $a<b \implies a^2 < b^2$, which is what you're asked to prove — so you can't use it. (A warning sign should be that nowhere you used $0 < a$.) Instead, you could, for instance, show that if $b-a>0$ (and $a>0$), then $b^2 - a^2 > 0$, without using a property like the ones you used above. (It's easy!) $\endgroup$
    – ShreevatsaR
    Jul 21, 2011 at 12:37
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    $\begingroup$ @Theo: It's even simpler to say that $b^2 - a^2 = (b-a)(b+a) > 0$ because both factors in the product are positive. :-) (The first factor because $b-a > 0$, and the second because both $a$ and $b$ are positive.) $\endgroup$
    – ShreevatsaR
    Jul 21, 2011 at 12:38
  • $\begingroup$ @ShrevatsaR: Agreed, much better :) $\endgroup$
    – t.b.
    Jul 21, 2011 at 12:48

5 Answers 5

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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.

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    $\begingroup$ Wonderful! I had thought I might need to show a progression from a < b to a^2 < b^2, but I had been going about it in the wrong way. Thank you for making it so simple and easy to understand. $\endgroup$
    – Bradford Fisher
    Jul 21, 2011 at 12:45
  • $\begingroup$ To be more precise, "monotonically increasing", since the result would not follow if it was monotonically decreasing. $\endgroup$
    – Nicholas Pipitone
    Jan 24, 2019 at 13:56
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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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    $\begingroup$ This is actually one of the methods that I tried. For whatever reason, however, I did not make the connection between aa < ab and ab < bb. Thank you for letting me know I was on the right path. Hopefully next time I will be more observative. $\endgroup$
    – Bradford Fisher
    Jul 21, 2011 at 12:55
  • $\begingroup$ Glad you found this answer useful. $\endgroup$
    – Shai Covo
    Jul 21, 2011 at 13:06
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$$0<a<b\Rightarrow 0\cdot a<a\cdot a<b\cdot a<b\cdot b\Rightarrow a^{2}<b^{2}.$$

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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 $b^2>a^2$.

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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).

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