>>13320526You are overcomplicating. Use the fact that $\sqrt a\geq0$ for any $a\geq0$. Then you just need to prove that each $x_i^2\geq0$. Here's how I would prove it
Lemma: $t^2\geq0$ for any $t\in\mathbb R$.
Proof: By trichotomy, either $t>0,t<0,$ or $t=0$. If $t>0$, then trivially $t^2>0$. If $t<0$, $-t>0$, so $(-t)^2>0$. However, $(-t)^2=(-1)^2t^2=t^2$, so $t^2>0$. If $t=0$, then $t^2=0$ Therefore, $t^2\geq0$.
Main proof:
By the lemma, $x_i^2\geq0$ for each $i$. Summing over $i$ gives $\sum_ix_i^2\geq0$. Since this sum is nonnegative, we can take the square root, so $\sqrt{\sum_ix_i^2}=|x|\geq0$ and we are done.
Also, word of advice about inequality proofs. When you're trying to prove an inequality, don't start from the inequality and work backwards; start from the given information and work forwards, ending with the inequality. Working backwards is not rigorous because you could have steps which aren't reversible. Think of it as a normal proof of anything else; in a normal proof we don't assume that the ending result is true and then "work backwards."
