Proof: $||u+v||^2≤(||u||+||v||)^2$ Using Cauchy Schwarz Inequality

Question

I can show that $||$u$+$v$||^2≤(||$u$||+||$v$||)^2$ with the same method shown in the solution.

To prove that $||$u$+$v$||≤||$u$||+||$v$||$, is it valid to just square root both sides since I have shown that $||$u$+$v$||^2≤(||$u$||+||$v$||)^2$ is true?

Answer 1

Suppose $X\subset \mathbb R$ and consider two elements of that set $a,b\in X$ and a function $f:X\to X$. If $f$ is weakly increasing then $a\ge b \Rightarrow f(a)\ge f(b)$.

In your case $X=\mathbb R_+$, $a=(||u||+||v||)^2$, $b=||u+v||^2$, and $f(x)=\sqrt{x}$. The square root is well defined in $\mathbb R_+$ and actually strictly increasing in it.