Proof of $x^p$ sub/super additive? What is the easiest proof that $x^p$ is sub (or super) additive for  $p&gt;1$ (or $p&lt;1$) and $p\in \mathbb R$.
This is the same as showing for  $a,b\geq 0$
$$a^p+b^p\leq (a+b)^p,$$
when $p\geq 1$. It is possible to take a logarithm and compare the Taylor series to prove the claim, but is there anything simpler?
Let $p&gt;1$. You can define for $x\geq 0$
$$f(x) = (a+x)^p\,;\qquad g(x) = a^p + x^p\,.$$
Clearly $f(0) = g(0)$. Now take the derivatives.
$$f'(x) = p(a+x)^{p-1}\,;\qquad g'(x)=p x^{p-1}\,.$$
Since $a&gt;0$ and $p&gt;1$ it's clear that for all positive $x$ you have $f'(x)\geq g'(x)$. This ends the proof for $p&gt;1$ since for all $x\geq 0$ you'll have $f(x)\geq g(x)$.
For $p&lt;1$ you can proceed in the same way and get the opposite inequality.
We have that the $p$-norm is a norm on $\mathbb{R}^2$ when $1\leqslant p &lt; +\infty$. This means that $\lVert.\rVert_p$ is subadditive. So let's define $a=(x,0)$ and $b=(0,y)$. The subadditivity means that
$$\lVert a + b \rVert = (|x|^p + |y|^p)^{1/p}\leqslant |x|+|y|=\lVert a \rVert + \lVert b \rVert$$
Which is exactly what we wanted:
$$|x|^p+|y|^p\leqslant (|x|+|y|)^p$$