Efficient Proof for all cases of triangle inequality

Question

The question asks Find an efficient proof for all the cases at once by first demonstrating

$$ (a+b)^2 \leq (|a|+|b|)^2 $$

My attempt at the proof:

for $a,b\in\mathbb{R}$

$$ \begin{align*} |a+b|^2 =&(a+b)^2 \text{ (Since $\forall x\in\mathbb{R}$, $x^2=|x|^2$)}\\ =& a^2+2ab+b^2\\ \leq&|a|^2+2|a||b|+|b|^2 \text{( Since $\forall x\in\mathbb{R}, x\leq |x|$)}\\ =&(|a| +|b|)^2 \end{align*} $$

But this is where I get stuck, I can arrive at the inequality but I do not know how to continue you from here. The question states that this should be efficient proof for all the cases, but jumping from that step to $|a+b|\leq |a|+|b|$ seems like a big jump with some steps missing. Any push in the right direction would be appreciated!

Answer 1

Note that for positive numbers $a$ and $b$, $$a^2 \le b^2 \implies a\le b $$ Thus $$|a+b|^2 \le (|a|+|b|)^2 \implies |a+b|\le|a|+|b|$$

Answer 2

All you need to say is that if $0<r^2<s^2$, then $r<s$. This follows from the fact that $r\geq s>0$ implies $r^2\geq s^2$.

Also, you are missing a square on $|a|$, and your last $\leq$ should be $=$.