Is the following generalization of Cauchy-Schwarz inequality true? 
Let $\{a_{1,i}\}_{i=1}^k,\{a_{2,i}\}_{i=1}^k,\dots ,\{a_{n_,i}\}_{i=1}^k$ be real sequences. Does the following inequality hold
$$(\sum_{i=1}^k a_{1,i}^2)\cdot(\sum_{i=1}^k a_{2,i}^2)\cdots(\sum_{i=1}^k a_{n,i}^2)\geq (\sum_{i=1}^k a_{1,i}a_{2,i}\cdots a_{n,i})^2$$
for all $k,n \in \mathbb N$?

It can be easily seen that this is the Cauchy-Schwarz inequality when $n=2$.
The motivation for the problem actually comes from the Cauchy-Schwarz inequality. While solving a Cauchy-Schwarz inequality problem, this problem came to my mind. I don't know if this is already a proved theorem in mathematics (because I am a high school student and I don't know much about inequalities). But I didn't find this on internet (I searched on google). So, I assume the problem statement is false. And a proof (or disproof) is needed for that.
My workings for $k=2$ and $n=3$:
However, I tried to prove the problem statement for $k=2$ and $n=3$ (and I think I actually proved that!). Here is my workings to do that:
For $a,b,c,d,e,f$ real numbers, we have from Cauchy-Schwarz inequality (which is for $n=2$ and $k=2$),
$$(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2$$
$$\implies (a^2+b^2)(c^2+d^2)(e^2+f^2) \geq (ac+bd)^2(e^2+f^2)$$
$$=(a^2c^2+2abcd+b^2d^2)(e^2+f^2)$$
$$=a^2c^2(e^2+f^2)+2abcd(e^2+f^2)+b^2d^2(e^2+f^2)$$
$$\geq a^2c^2e^2+2abcdef+b^2d^2f^2$$
$$=(ace+bdf)^2$$
as desired.

I hope my workings are correct. So, I have the following questions:

*

*Is the firstly stated problem statement true? If it is, how to prove that?

*If it is not true, are there some other values (like $k=2$ and $n=3$ as in the above) for which the statement is true?

Any help would be appreciated and please try to answer the questions so that a high school student can understand them (if it is not possible, then no problem).
 A: Here is a bit of context (that is, admittedly, overkill, see the last paragraph of my answer). The inequality that you mention is true and it is a special case of the generalized Hölder inequality. More precisely, let $a_1,a_2,\dots,a_n\in\ell^{n}$ for some given integer $n\ge2$. Then generalized Hölder tells you that
$$\lVert a_1\cdot a_2\cdot\ldots\cdot a_n\rVert_{\ell^1}\le\lVert a_1\rVert_{\ell^n}\cdot\ldots\cdot\lVert a_n\rVert_{\ell^n}.$$
This tells you that
$$\lVert a_1\rVert_{\ell^n}\cdot\ldots\cdot\lVert a_n\rVert_{\ell^n}\le\lVert a_1\rVert_{\ell^2}\cdot\ldots\cdot\lVert a_n\rVert_{\ell^2}.$$
Your case, vectors in $\mathbb R^k$, are a special case of the above, since a vector $a=(a^{(1)},\dots,a^{(k)})\in\mathbb R^k$ can always be embedded into $\ell^n$ as the sequence $$(a^{(1)},\dots,a^{(k)},0,0,\dots).$$

Finally I want to apologize for using a concept, the $\ell^p$ spaces (see for instance https://en.wikipedia.org/wiki/Lp_space#The_p-norm_in_finite_dimensions and the following section), that are definitely not encountered in high school .
A: This proofs assume you are familiar with the concept of mathematical induction.
Statement does not hold for $n=1, k \neq 1$.
For every $k$, proceed by induction on $n$ (case $n=2$ is Cauchy-Schwarz). Then, we can reduce to the case $n-1$ as
$\left( \sum_{i=1}^k a_i^2 \right) \left( \sum_{i=1}^k b_i^2 \right) \geq \sum_{i=1}^k (a_ib_i)^2$
This inequality holds because $(a_ib_j)^2 \geq 0; i \neq j$. Further details remain as an execise.
A: Assume w.l.o.g. that all the numbers $\{a_{i,j}\}$ are nonnegative. The case $n=1,2$ is clear, so assume as induction hypothesis that
$$
\sum_{j=1}^ka_{1,j}\cdots a_{(n-1),j} \leq \left(\sum_{j=1}^ka_{1,j}^2\right)^{1/2}\cdots\left(\sum_{j=1}^ka_{(n-1),j}^2\right)^{1/2}.
$$
Define $c_j = a_{n,j}/\left(\sum_{j=1}^ka_{n,j}^2\right)^{1/2}$, then using the hypothesis:
\begin{align}
\sum_{j=1}^kc_ja_{1,j}\cdots a_{(n-1),j} &= \sum_{j=1}^kc_j^{1/(n-1)}a_{1,j}\cdots c_j^{1/(n-1)}a_{(n-1),j} \leq\left(\sum_{j=1}^kc_ja_{1,j}^2\right)^{1/2}\cdots\left(\sum_{j=1}^kc_ja_{(n-1),j}^2\right)^{1/2} \\&\leq\left(\sum_{j=1}^ka_{1,j}^2\right)^{1/2}\cdots\left(\sum_{j=1}^ka_{(n-1),j}^2\right)^{1/2},
\end{align}
since $c_j \leq 1$ for all $j$. Substituting back $a_{n,j}$, and multiplying both sides by $\left(\sum_{j=1}^ka_{n,j}^2\right)^{1/2}$:
$$
\sum_{j=1}^ka_{1,j}\cdots a_{n,j} \leq\left(\sum_{j=1}^ka_{1,j}^2\right)^{1/2}\cdots\left(\sum_{j=1}^ka_{n,j}^2\right)^{1/2},
$$
thus the target inequality is proven by induction.
EDIT: Another silly proof. Let $\{x_{i,j}\}$ be nonnegative and assume that $n \geq 2$, then using AM-GM
$$
x_{1,j}\cdots x_{n,j}\leq \frac{x_{1,j}^n+\cdots+x_{n,j}^n}{n}.
$$
Now, make the substitution $x_{i,j} = a_{i,j}/\left(\sum_{l=1}^k a_{i,l}^2\right)^{1/2}$, then since $0\leq x_{i,j} \leq 1$
$$
x_{1,j}\cdots x_{n,j}\leq \frac{x_{1,j}^n+\cdots+x_{n,j}^n}{n}\leq\frac{x_{1,j}^2+\cdots+x_{n,j}^2}{n}.
$$
Summing both sides along $j$ gives:
$$
\frac{\sum_{j=1}^ka_{1,j}\cdots a_{n,j}}{\left(\sum_{j=1}^ka_{1,j}^2\right)^{1/2}\cdots\left(\sum_{j=1}^ka_{n,j}^2\right)^{1/2}} \leq 1,
$$
which is the target inequality.
A: I think your proof for the case $k = 2$ and $n = 3$ is valid.
Without explicitly using mathematical induction, as in Jorge's
answer - although induction is always finally needed to justify an
informal proof like this - one can see that the inequality for
general $n \geqslant 2$ follows almost immediately from Cauchy's
inequality, simply by losing most of the terms from the expanded
product of the last $n - 1$ bracketed sums, thus:
\begin{multline*}
\left(\sum_{i=1}^ka_{1,i}^2\right)
\left(\sum_{i=1}^ka_{2,i}^2\right) \cdots
\left(\sum_{i=1}^ka_{n,i}^2\right)
\geqslant
\left(\sum_{i=1}^ka_{1,i}^2\right)
\left(\sum_{i=1}^ka_{2,i}^2 \cdots a_{n,i}^2\right) = \\
\left(\sum_{i=1}^ka_{1,i}^2\right)
\left(\sum_{i=1}^k(a_{2,i} \cdots a_{n,i})^2\right)
\geqslant
\left(\sum_{i=1}^ka_{1,i}(a_{2,i} \cdots a_{n,i})\right)^2 =
\left(\sum_{i=1}^ka_{1,i}a_{2,i} \cdots a_{n,i}\right)^2.
\end{multline*}
This proof "gives away" so much that the resulting inequality,
when $n > 2,$ is very weak. This is illustrated by the fact that if
there are $b_1, b_2, \ldots, b_n$ such that $a_{j,i} = b_j,$ for
$j = 1, 2, \ldots, n,$ and $i = 1, 2, \ldots, k,$  then the
inequality reduces to
$(kb_1^2)(kb_2^2)\cdots(kb_n^2) \geqslant (kb_1b_2 \cdots b_n)^2,$
i.e., $k^n \geqslant k^2,$ which is of little interest when $n > 2$!
That probably explains why the case $n > 2$ is seldom mentioned.  I
did find the case $n = 3$ given as Exercise XVa, problem 37 in
Clement V. Durell, Advanced Algebra, Vol. III
(Bell, London 1937). A more up-to-date reference is Exercise 1.3 in
J. Michael Steele, The Cauchy-Schwarz Master Class
(Cambridge University Press / Mathematical Association of America
2004). Steele gives a surprisingly complicated proof, which is why I
thought it worth giving this very simple one. (In essence it
duplicates Jorge's proof, but the idea seems worth repeating in
different words.)
