TITLE: For vector p-norm, can we prove it is decreasing without using derivative?
QUESTION [6 upvotes]: For vector p-norm defined as $(∑_{i=1}^n x_i^p )^{\frac{1}{p}}$ for any $p\ge 1$ and vector ${\bf{x}}=\{x_1,...,x_n\}$. The following proves it is decreasing with respect to $p$ by taking derivative (you don't need to read the whole proof, just have a look),



However, I am thinking if there is another approach without using the derivative. Is there any proof for monotonicity of p-norm without using derivatives? The upper bound can be proved by Holder's inequality by Relations between p norms
We have plenty of inequalities that lead to the definition of p-norm: Young's inequality, Jensen's inequality, Holder's inequality, Minkowski’s inequality. Maybe there is a proof using those inequalities?

REPLY [3 votes]: In Help show $(∑_{i=1}^n |x_i | )^p≥∑_{i=1}^n |x_i |^p $ using common inequalities (like Holder's inequality) you find a proof for the special case
$$\|x\|_1 \ge \|x\|_p.$$
Now, replace $x$ by $|y|^r$ and you find
$$\|y\|_r^{1/r}=\||y|^r\|_1 \ge \||y|^r\|_p = \|y\|_{rp}^{1/r}.$$
Taking the $r$th root you have
$$\|y\|_r \ge \|y\|_{r p}$$
and this settles the general case.