Triangle inequality in $L^p$ space for $p\in (0,1)$. For the space $L^p$ with $p\in (0,1)$, I know that there is $K>1$ s.t. $$\|x+y\|_{L^p}\leq K(\|x\|_{L^p}+\|y\|_{L^p}).$$

Q1) What is the minimal $K$ ? Or at least, can we take $K=2^{p-1}$ ?

Q2) Now, do we have a constant $C$ s.t. $$\|x-y\|_{L^p}\geq C|\|x\|_{L^p}-\|y\|_{L^p})\ \ ?$$
Attempt
$$\|x\|_{L^p}\leq K\|x-y\|_{L^p}+K\|y\|_{L^p}\implies \frac{1}{K}\|x\|_{L^p}-\|y\|_{L^p}\leq \|x-y\|_{L^p}.$$
The problem is neither $\frac{1}{K}>1$ nor $-1>\frac{-1}{K}$, so I cant conclude...
 A: For $0<p<1$, we have $(a+b)^p \leq a^p+b^p$ for positive $a$ and $b$:
$$ (a+b)^p-a^p = \int_0^{b} \frac{p}{(a+t)^{1-p}} \, dt \leq \int_0^{b} \frac{p}{t^{1-p}} \, dt = b^p, $$
since $\frac{1}{a+t}\leq \frac{1}{t}$. Integrating this gives
$$ \lVert x+y \rVert_p^p \leq \lVert x\rVert_p^p+\lVert y\rVert_p^p, $$
so it now suffices to find a constant so that for $a,b>0$,
$$ a^p+b^p \leq K^p(a+b)^p. $$
For $0<p \leq 1$, $t \mapsto t^p$ is concave, so
$$ \left(\frac{a+b}{2}\right)^p \geq \frac{1}{2}(a^p+b^p), $$
or
$$ a^p+b^p \leq 2^{1-p}(a+b)^p. $$
Therefore,
$$ \lVert x+y \rVert_p \leq (\lVert x\rVert_p^p+\lVert y\rVert_p^p)^{1/p} \leq 2^{1/p-1} (\lVert x\rVert_p+\lVert y\rVert_p), $$
so $K=2^{1/p-1}$. (Notice that $K=2^{p-1}$ is no good since it is smaller than $1$ when $0<p<1$.)

The answer to the second part should be no, since $K>1$ is not good enough to carry out the usual proof. Writing $x=y+z$, we want to show there is no $C$ so that
$$\lVert z \rVert_p \geq C \big\lvert \lVert y+z \rVert_p - \lVert y \rVert_p \big\rvert $$
for every $z$; equivalently, to show that there are $z$ for which
$$ \frac{\lVert y+tz \rVert_p - \lVert y \rVert_p}{t}, $$
diverges as $t \to 0$, since then no $C$ will suffice. A simple example is given by points in $\mathbb{R}^2$: let $y=(1,0)$, $z=(0,1)$. Then
$$ \lVert y+z \rVert_p - \lVert y \rVert_p = (1+\lvert t \rvert^p)^{1/p} - 1 = \frac{\lvert t \rvert^p}{p} + O(t^{2p}), $$
and this diverges as $t \to 0$. (This also makes plain why $p<1$ is significantly different from $p \geq 1$.)
