TITLE: best constant in quasi triangle inequality for $L^p$ spaces with $0 < p \le 1$
QUESTION [5 upvotes]: Currently doing a problem that ask me to prove the best constant $C$ (depending on $p$) such that the quasi triangle inequality $||f+ g||_p \le C (||f||_p + ||g||_p)$ for $L^p $ spaces holds, where $0< p \le 1$, is $2^{1/p -1 } $. 
My approach is I already proved $||f+g||_p^p \le ||f||_p^p + ||g|| _p ^p$. So it will be suffices to show that $(||f||_p^p + ||g|| _p^p ) ^{1/p} \le 2^{1/p -1} ( ||f||_p + ||g||_p)$. But I have no idea how to proceed this. Any help is appreciated.

REPLY [7 votes]: I think i got it!
By what I have proved I have $ ||f+g||_p \le ( ||f||_p^p + ||g||_p^p) ^{1/p} $. Then since $ ||f||^p_p , ||g||^p _p \ge 0$ and $x \mapsto x^{1/p}$ is convex when $ 0 < p \le 1$ we have by Jensen's inequality $ (2 ( \frac{1}{2} ||f||^p_p + \frac{1}{2} ||g||^p_p))^{1/p} \le 2^{1/p} ( \frac{1}{2} ||f||_p + \frac{1}{2} ||g||_p ) = 2^{1/p - 1} (||f||_p + ||g||_p)$. Thus we get $||f+g||_p \le 2^{1/p - 1} (||f||_p + ||g||_p)$ as desired.
To see this is the best constant, we take $f = 1_{[-1,0]}$ and $g = 1_{[0,1]}$, then $||f+g||_p = 2^{1/p} =  2^{1/p - 1} (||f||_p + ||g||_p)$. Actually any two functions $f,g \in L^p$ that have disjoint support will achieve the equality.