For all $f,g\ |f+g|_p\leq |f|_p+|g|_p$ then $p\geq 1.$

Question

Consider $L^p(\mathbb{R}^d$) when $0<p<\infty$. Show that if for all $f,g,\ ||f+g||_p\leq ||f||_p+||g||_p$ then $p\geq 1$. Hint. If $p<1$ then for all $x,y>0$ then $(x+y)^p<x^p+y^p$.

I dont understand the hint. If $p<1$, with $f(x)=(x+y)^p-x^p+y^p$, $f'(x)<0$ then $f$ decreacing. Therefore $f(0)<f(x)$ then $(x+y)^p<x^p+y^p$. But, why $(x+y)^p<x^p+y^p$ proves the affirmation?

Answer 1

Let $A,B$ be disjoint sets each with measure $1$ and $f=aI_A, g=bI_B$. Then $|f+g|_p=(\int |f+g|^{p})^{1/p} =(|a|^{p}+|b|^{p})^{1/p}$ and $|f|_p=|a|, |g|_p\|b|$, hence the given inequality gives $(|a|^{p}+|b|^{p})^{1/p} \leq |a|+|b|$ or $|a|^{p}+|b|^{p} \leq (|a|+|b|)^{p}$ which is a contradiction to the inequality you have proved.