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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!

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    It is a well known generalization of Holder's Inequality (provable by straightforward induction) that $$\left|\left|\prod_{j=1}^k f_j\right|\right|_r \leq \prod_{j=1}^k\left|\left|f_j\right|\right|_{p_j},$$ where $$\sum_{j=1}^k\frac{1}{p_j} = \frac{1}{r},$$ for $r \in (0,\infty), p_1,\ldots,p_k \in (0,\infty],$ and $f_j \in L^{p_j}$. In your case, take $r =1$ and $p_j = k$. – 2017-01-30
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    I did think if such a thing is true but I could not make sense of this LHS, $\prod_j f_j$. If $f_j \in L^{p_j}$ then $f_j$ is a vector. Then what does this LHS mean? – 2017-01-30
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    The $f$'s are more appropriately thought of as vector-valued functions taking values in the $n$-dimensional reals. You can then think about a generalization (to arbitrary $k$) of the product measure version of Holder's inequality with the counting measure. In essence, this amounts to the product being treated like a Hadamard product if you think about the $f$'s as being vectors. – 2017-01-30
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    This and much else is captured by the Brescamp-Lieb inequalities https://en.wikipedia.org/wiki/Brascamp%E2%80%93Lieb_inequality – 2017-02-08

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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.