Is the following generalization of Cauchy-Schwarz inequality true?

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.)

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 😅.

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.