<s>[INST] Simplification of proof for $\exists c>0$ such that $|x+y|\leq c(1+\sqrt{x^2+y^2})$ Question: Consider function $f(x,y)=|x+y|$ for $x,y\in \mathbb{R}$. Show that $\exists c>0$ such that $|x+y|\leq c(1+\sqrt{x^2+y^2})$.
My proof:
i) if $|x|,|y|\leq 1$, then $|x+y|\leq 2\leq 2(1+\sqrt{x^2+y^2})$;
ii): if $|x|\leq 1$ and $|y|\geq 1$, then $|x+y|\leq 1+|y|\leq 1+\sqrt{x^2+y^2}$;
iii): if $|x|,|y|\geq 1$ such that $|x|\geq |y|$, then $|x+y|\leq2|x|\leq2(1+\sqrt{x^2})\leq 2(1+\sqrt{x^2+y^2})$.
Hence, we conclude, and $c=2$ works for the claim.
However, I feel like there should be a simpler proof which does not prove in a case-by-case way like I did.
 [/INST] Notice that
$$|x|=\sqrt{x^2}\leq\sqrt{x^2+y^2}\quad\text{and}\quad|y|=\sqrt{y^2}\leq\sqrt{x^2+y^2}$$
and so
$$|x+y|\leq|x|+|y|\leq 2\sqrt{x^2+y^2}<2\left(1+\sqrt{x^2+y^2}\right)$$
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