Understanding why Minkowski’s inequality doesn't hold true for $0 < p < 1$? The triangle inequality given by
$\left(\sum_{i=1}^n |x_i+y_i|^p\right)^{1/p}\leq \left(\sum_{i=1}^n |x_i|^p\right)^{1/p} + \left(\sum_{i=1}^n |y_i|^p\right)^{1/p}$
is known as “Minkowski’s inequality" which holds true for $1\leq p <\infty$ while for $0 < p < 1$ it doesn't  hold. 
Searching through net I have found that  this inequality holds for $1\leq p <\infty$  is related to the observation that for such $p$ the function $x \to x^p$ for $x \geq 0$ is convex.
The failure of the triangle inequality is related to the observation that for $0 <
p < 1$,the function $x \to x^p$ for $x \geq 0$ is not convex.
I am not able to understand these points. Could anybody explain me?
Thanks
 A: A function $h(x)$ is convex if it satisfies
$$h\left ( \alpha x_1 + (1-\alpha)x_2 \right) \le \alpha h(x_1) + (1-\alpha)h(x_2),
\quad\alpha\in [0,1]\tag{1}$$
Geometrically this means that for $x_1\le x\le x_2$ the graph of $h(x)$ must lie under the line connecting the points $h(x_1)$ and $h(x_2)$. Inspecting the graph of $h(x)=x^p$ it is easy to see that the function $x^p$ is convex for $p\ge 1$, but not for $0<p<1$.
In the proof of Minkowski's inequality they make use of exactly this fact. The proof starts by proving that the sum of some functions $f$ and $g$ has finite $p$-norm if both $f$ and $g$ do. We get
$$\left | \frac{f}{2}+\frac{g}{2}\right |^p\le
\left | \frac{|f|}{2}+\frac{|g|}{2}\right |^p$$
This inequality has nothing to do with convexity but only follows from taking the magnitudes of $f$ and $g$. In the following step, however, we make use of convexity:
$$\left | \frac{|f|}{2}+\frac{|g|}{2}\right |^p\le
\frac{|f|^p}{2}+\frac{|g|^p}{2}$$
Here we made use of (1) with $\alpha=1/2$ and $h(x)=x^p$, $p\ge 1$.
The rest of the proof builds on this result, that's why convexity (and therefore $p\ge 1$) is necessary for Minkowsi's inequality to hold.
A: I have a good(well, I think it's good) interpretation. The Minkowski's inequality simply suggests that $L^p$ is a normed space. But you can show that $L^p$ isn't even locally convex space if $0<p<1$.
