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 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.