Is this some kind of Holder's inequality?

Question

I am seeing this particular inequality for an array of real numbers $\{ x_{ij} \}$,

$$\sum_i \vert \prod_{j=1}^k x_{ij} \vert \leq \prod_{j=1}^k ( \sum_i \vert x_{ij}\vert ^k)^{\frac{1}{k}}$$

I would like to know the proof of this!

Answer 1

Proof by induction on $k$. When $k = 1$ the inequality above is an equality. Now suppose $k > 1$ and the result holds for all positive integers less than $k$. By Hölder's inequality (with conjugate exponents $k$ and $k/(k-1)$),

$$\sum_{i} \left\lvert \prod_{j = 1}^k x_{ij}\right\rvert = \sum_i \left\lvert \prod_{j = 1}^{k-1} x_{ij}\right\rvert \lvert x_{ik}\rvert \le \left(\sum_i \left\lvert \prod_{j = 1}^{k-1} x_{ij}\right\rvert^{k/(k-1)}\right)^{(k-1)/k}\left(\sum_i \lvert x_{ik}\rvert^k\right)^{1/k}.$$

Now

$$\left(\sum_i \left\lvert \prod_{j = 1}^{k-1} x_{ij}\right\rvert^{k/(k-1)}\right)^{(k-1)/k} = \left(\sum_i \left\lvert \prod_{j = 1}^{k-1} x_{ij}^{k/(k-1)}\right\rvert\right)^{(k-1)/k} \le \prod_{j = 1}^{k-1} \left(\sum_i \lvert x_{ij}^{k/(k-1)}\rvert^{k-1}\right)^{1/k},$$

using the induction hypothesis in the last step. Thus

$$\left(\sum_i \left\lvert \prod_{j = 1}^{k-1} x_{ij}\right\rvert^{k/(k-1)}\right)^{(k-1)/k} \le \sum_{j = 1}^{k-1} \left(\sum_i \lvert x_{ij} \rvert^k\right)^{1/k},$$

and consequently

$$\sum_i \left\lvert \prod_{j = 1}^k x_{ij}\right\rvert \le \prod_{j = 1}^{k-1} \left(\sum_i \lvert x_{ij}\rvert^k\right)^{1/k} \left(\sum_i \lvert x_{ik}\rvert^k\right)^{1/k} = \prod_{j = 1}^k \left(\sum_i \lvert x_{ij}\rvert^k\right)^{1/k},$$

as desired.