Minkowski's integral inequality in case $0Let $(X,\mu)$ be a measure space, $L^p(X,\mu)$ be the set of all complex-valued $\mu-$measurable functions on $X$ whose modulus to the $p$th power is integrable. I know how to prove the Minkowski's inequality for $1\leq p\leq \infty$. In case $0<p<1$, we have the following: $|f+g|_{L^p(X,\mu)}\leq2^{(1-p)/p}\left(|f|_{L^p(X,\mu)}+g_{L^p(X,\mu)}\right)$ for any $f,g\in {L^p(X,\mu)}$ and $0<p<1$.
My question is how to prove the last inequality?.
 A: Many $L^p$ estimates follow from the convexity (p>1) and concavity (p<1) of $x^p$. 
Indeed, since $0<p\le 1$, the mapping $x\mapsto x^p$ is concave on $\mathbb{R}_{\ge 0}$. Moreover, since $0\mapsto 0^p=0$, it is a good exercise to show that $x\mapsto x^p$ is subadditive (solution). Thus we derive the pointwise bound $|f(x)+g(x)|^p\le |f(x)|^p+|g(x)|^p$. Consequently, $||f+g||_p^p\le ||f||_p^p+||g||_p^p$. 
Again, employing the concavity of $x\mapsto x^p$, we have that 
$\begin{align*} ||f+g||_p^p&\le ||f||_p^p+||g||_p^p\\ &=2\left(\frac{1}{2}||f||_p^p+\frac{1}{2}||g||_p^p\right)\\ &\le 2\left(\frac{1}{2}||f||_p+\frac{1}{2}||g||_p\right)^p\\ &=2^{1-p}(||f||_p+||g||_p)^p\end{align*}$
Now take $p$-th roots of both sides. As an aside, I should mention that this constant is best (when considering $L^p(X,\mathcal{M},\mu)$ over all measure spaces). Indeed, to see that $2^{(1-p)/p}$ is optimal, it suffices to exhibit a specific measure space and pair of functions to witness it. Luckily we wont have to look very far. Just take $\mathbb{R}$ with the Borel $\sigma$-algebra with Lebesgue measure. 
Let $f=\chi_{[0,1]}$ and $g=\chi_{[1,2]}$. Then $||f+g||_{p}=2^{1/p}=2^{1/p-1}2=2^{1/p-1}(||f||_p+||g||_p)$. 
