Squaring both sides of an inequality: attempt to prove a general rule

Question

I have attempted to produce a proof of the intuitive rule for squaring inequalities, according to which, given any two numbers x and y and regardless of their sign,

1) if |x| < |y| then $x^2<y^2$;

2) if |x| > |y| then $x^2>y^2$.

Proof for Case 1:

Let x and y being any two real numbers such that |x| < |y|.

• Multiplying both sides by |x| gives |x||x| < |y||x|. Hence, by the result |a||b| = |ab|, we have |x|$^2$<|xy|;

• Multiplying both sides by |y| gives |y||x| < |y||y|. Hence, by the same result, we get |xy|<|y|$^2$.

Now, merging the two inequalities we get |x|$^2$<|xy|<|y|$^2$, to give |x|$^2$<|y|$^2$. By the result |a|$^2$ = _a_$^2$ we get _x_$^2$<_y_$^2$. This completes the proof in the first case.

Proof for Case 2:

Let x and y being any two real numbers such that |x| > |y|.

• Multiplying both sides by |x| gives |x||x| > |y||x|. Hence, by the result |a||b| = |ab|, we have |x|$^2$>|xy|;

• Multiplying both sides by |y| gives |y||x| > |y||y|. Hence, by the same result, we get |xy|>|y|$^2$.

Merging the two inequalities we get |x|$^2$>|xy|>|y|$^2$, to give |x|$^2$>|y|$^2$. By the result |a|$^2$ = _a_$^2$ we get _x_$^2$>_y_$^2$. This completes the proof in the second case.

Substantially this proof (hopefully correct) follows the same path as the proof for squaring both sides of an inequality. The difference is that here we take into account only absolute values, which avoids us the tricky task of considering signs.

Any comment or suggestion to improve (or disprove!) this would be most welcome!

Answer 1

Your proof seems correct, although a bit too complicated.

If you know that $x \mapsto x^2$ is a strictly increasing function on $\Bbb{R}_{\geq 0}$, then you immediately know that $|x| > |y|$ iff $x^2 > y^2$ for every $x,y \geq 0$ (by definition of strictly increasing). To prove that, the quickest way would be to observe that $\frac{d}{dx}(x^2) = 2x \geq 0$ for $x \geq 0$, with equality only for $x=0$.

---

To keep things elementary, you could instead observe that $|x| > |y|$ implies $$ |x||y| < |x||x| = x^2 \qquad \text{and} \qquad |x||y| > |y||y| = y^2 $$ Putting the two together gives $|x| > |y| \Rightarrow x^2 > y^2$. Reversing the roles of $x$ and $y$ gives (1), but note that the two cases are really just the same thing.

Answer 2

Notice that both inequalities follow from proving that

$\;\;\;$If $0\le a<b,$ then $a^2<b^2$; and

$\;\;\;0\le a<b\implies a^2\le ab\;$ and $\;ab<b^2\implies a^2<b^2$.