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?
 A: 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 $a^2 < b^2$.
A: 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.
A: 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.$$\begin{align} 0 \le a \lt b \implies b-a>0 \\ \implies (b-a)^2>0 \\ \implies b^2-2ab+a^2>0 \\ \implies 2ab<a^2+b^2 \\ \implies 2a(a)<a^2+b^2 \quad  \text{by assumption that} \quad a<b \\ \implies 2a^2<a^2+b^2 \\ \implies a^2<b^2 \end{align}$$
A: 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.
A: If $a=0$, then it is trivial since then $b^2>0=a$
Assume $0<a<b$ then:
$aa<ba<bb$.
A: $$a^2-b^2 =\underbrace{(a-b)}_{<0}\underbrace{(a+b)}_{>0} <0 \implies a^2< b^2$$
A: $b^2 - a^2 \geq b^2 - ba$ 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: $b^2 - a^2 \geq b^2 - ba = b(b-a) > 0$.
