Help showing the triangle inequality holds for a particular metric on $\mathbb{R}^n$ Given $0<p<1$, and $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ in $\mathbb{R}^n$, I am trying to show that the function defined by 
$$d_p(x,y)=\sum_{j=1}^{n}{|x_j-y_j|}^p$$
satisfies the triangle inequality so that I can verify it is a metric. 
So far, I haven't had any luck. Thanks for any help. 
 A: We begin with the case $n=1$. Looking at the graph of $t\mapsto t^p$ $(t\geq0)$ when $0<p<1$ we see that for $t\geq0$, $s\geq0$ one has
$$(t+s)^p-t^p\leq s^p-0^p\ ,$$
and this implies $(t+s)^p\leq t^p+s^p$. For given real $x$, $y$, $z$ put $t:=|x-y|$, $\>s:=|y-z|$. We then have $|x-z|\leq t+s$ and therefore $$d(x,z):=|x-z|^p\leq(t+s)^p\leq t^p+s^p=d(x,y)+d(y,z)\ .$$
When $n>1$ we therefore get
$$d(x,z):=\sum_{k=1}^n d(x_k,z_k)\leq\sum_{k=1}^n\bigl(d(x_k,y_k)+d(y_k,z_k)\bigr)=d(x,y)+d(y,z)\ .$$
A: We will prove each of the following (each of which depends on the previous):


*

*for $c > 0$ the function $g(x) = x^c$ is increasing on $[0,\infty)$;

*for $\alpha > 0$ and $\beta > 1$ the function $f(x) = (\alpha + x)^\beta - x^\beta$ is increasing on $[0,\infty)$;

*for $a,b\geq 0$ and $\beta > 1$, $(a+b)^\beta \geq a^\beta + b^\beta$;

*for $a,b \in \mathbb{R}$ and $\beta > 1$, $|a+b| \leq (|a|^{\frac{1}{\beta}} + |b|^{\frac{1}{\beta}})^\beta$;

*for $a,b \in \mathbb{R}$ and $0<p<1$, $|a+b|^p \leq |a|^{p} + |b|^{p}$


(To view any of proofs move your mouse over them)
Proof of 1:

 This follows from the fact that $g'(x) = cx^{c-1}>0$ for $x>0$.

Proof of 2:

 Differentiating $f$ in the second statement we obtain $f'(x)=\beta\left[(\alpha+x)^{\beta-1} - x^{\beta-1}\right]$ which is positive for $x>0$ since $g$ is increasing and $\alpha +x > x$.

Proof of 3:

 For $a,b>0$ setting $\alpha=a$ from 2. we obtain that $(a+b)^\beta - b^\beta=f(b)>f(0)=a^\beta$ and hence $(a+b)^\beta > a^\beta + b^\beta$. For either $a=0$ or $b=0$ the inequality obviously holds.

Proof of 4:

 By the triangle inequality we obtain $|a+b| \leq |a| + |b|$. Applying 3. to $|a|^{\frac{1}{\beta}}$ and $|b|^{\frac{1}{\beta}}$ we obtain $(|a|^{\frac{1}{\beta}} + |b|^{\frac{1}{\beta}})^\beta \geq |a| + |b|$. Combining these we obtain the desired identity.

Proof of 5:

 Setting $\beta=\frac{1}{p}$ in 4. we obtain  $|a+b| \leq (|a|^{p} + |b|^{p})^{\frac{1}{p}}$. By 1. we have that $|a+b|^p \leq |a|^{p} + |b|^{p}$

