<s>[INST] If $0 \le a \lt b$ prove that $a^2 < b^2$ Below is how I prove it.
Case 1: $a = 0$


*

*$0^2 < b^2$ where $b$ is a positive number.

*$0 < b \times b$

*A positive number times a positive number is always positive.

*It is true.


Case 2: $a > 0$


*

*$a < b \Rightarrow a + x = b$

*$a^2 < b^2 \Rightarrow a^2 < (a+x)^2$

*$a^2 < (a+x) \times (a+x)$

*$a^2 < a^2 + 2ax + x^2$

*$0 < 2ax + x^2$

*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?
 [/INST] 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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