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Below is how I prove it.

Case 1: $a = 0$

  1. $0^2 < b^2$ where $b$ is a positive number.
  2. $0 < b \times b$
  3. A positive number times a positive number is always positive.
  4. It is true.

Case 2: $a > 0$

  1. $a < b \Rightarrow a + x = b$
  2. $a^2 < b^2 \Rightarrow a^2 < (a+x)^2$
  3. $a^2 < (a+x) \times (a+x)$
  4. $a^2 < a^2 + 2ax + x^2$
  5. $0 < 2ax + x^2$
  6. It is true because $a, x$ are positive numbers.

I was wondering a) if my prove is correct and b) if there are other straightforward way to prove this?

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  • $\begingroup$ You could just say that $x\mapsto x^{2}$ is a non-decreasing function on $\mathbb{R}_{+}$, so preserves the inequality. $\endgroup$ – Nicolas Feb 17 '15 at 16:26
  • $\begingroup$ See here for a simpler proof of a slightly stronger result. $\endgroup$ – Bill Dubuque Feb 17 '15 at 16:26
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You don't need cases. Just say:

$a^2 - b^2 = (a - b)(a + b)$.

Since $a \ge 0$ and $ b>0$, $a + b > 0$ and $a - b < 0$ since $a < b$, thus: $(a -b)(a +b) < 0$. So $a^2 - b^2 < 0$. Thus $a^2 < b^2$.

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If both are positive, then: $$a < b \implies \begin{cases} a^2 \leq ab \\ ab < b^2\end{cases} \implies a^2<b^2,$$ where we use transitivity in the last step.

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    $\begingroup$ This is neat, but if $0\le a$ and $a\lt b$ you can multiply by $a$ to obtain $a^2\le ab$, which is all you need, since the other inequality is strict. This avoids dealing with two cases. $\endgroup$ – Mark Bennet Feb 17 '15 at 16:32
  • $\begingroup$ Good point. I edited it in, thanks! :) $\endgroup$ – Ivo Terek Feb 17 '15 at 16:33
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Your proof of the case for $a=0$ is correct. Your case for the proof of $a>0$ is incorrect, because on the second line you assume what you are trying to show. Here is a way to do away with cases.$$\begin{align} 0 \le a \lt b \implies b-a>0 \\ \implies (b-a)^2>0 \\ \implies b^2-2ab+a^2>0 \\ \implies 2ab<a^2+b^2 \\ \implies 2a(a)<a^2+b^2 \quad \text{by assumption that} \quad a<b \\ \implies 2a^2<a^2+b^2 \\ \implies a^2<b^2 \end{align}$$

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  • $\begingroup$ So, I cannot assume what I try to show can work backward, right? $\endgroup$ – idonno Feb 17 '15 at 16:43
  • $\begingroup$ You are right. Otherwise, you are assuming what you need to show is true, to prove that what you need to show is true. This is circular reasoning :) But, you could assume the opposite of what you want to show for a proof by contradiction. That is, you suppose for the sake of contradiction that $a^2 \geq b^2$ and show that contradicts the assumption that $0 \leq a <b$. $\endgroup$ – graydad Feb 17 '15 at 16:46
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Hint: You could use a Calculus approach.

Let $f: [0,+\infty) \to \mathbb R $ defined by $f(x) = x^2$ and notice that $f'(x) = 2x \geq 0$. The function $f$ is non-decreasing.

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If $a=0$, then it is trivial since then $b^2>0=a$

Assume $0<a<b$ then:

$aa<ba<bb$.

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$$a^2-b^2 =\underbrace{(a-b)}_{<0}\underbrace{(a+b)}_{>0} <0 \implies a^2< b^2$$

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$b^2 - a^2 \geq b^2 - ba$ since $a^2 \geq ba$. ("We're subtracting something bigger, so the total becomes smaller")

$b^2 - ba = b(b-a) > 0$ as each factor is positive.

Putting it all together: $b^2 - a^2 \geq b^2 - ba = b(b-a) > 0$.

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