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).

  • 1
    $\begingroup$ The notations $a_{1_i}$, $a_{2_i}$ and $a_{n_i}$ don't make sense - you meant $a_{1,i}$, etc. $\endgroup$ Jul 3, 2021 at 15:48
  • 1
    $\begingroup$ the difference is that $a_{1,i}$ is standard notation for a doubly-indexed sequence, while $a_{1_i}$ is simply meaningless. It does not mean anything, because it doesn't have a definition. What is "$1_i$"??? $\endgroup$ Jul 3, 2021 at 15:59
  • 1
    $\begingroup$ maybe it's still not clear. sequence notation is shorthand for functional notation: $x_j=f(j)$. So when you write $a_{n_1}$ that should mean $a(n_i)$; $a$ is a function of one variable, evaluated at the point $n_1$. That makes sense (as opposed to the $1_i$ which is simply meaningless), but it's not what you meant. Its a double sequence, so a function of two variables: you meant $a(n,1)$, also written $a_{n,1}$. $\endgroup$ Jul 3, 2021 at 16:05
  • 1
    $\begingroup$ well done for the proof. In the last passage you have a typo $\endgroup$
    – Thomas
    Jul 3, 2021 at 20:15
  • $\begingroup$ There is a generalisation of the Cauchy-Schwarz inequality to a $k \times n$ matrix of real numbers $(a_{j,i})_{1\leqslant i\leqslant k,\ 1\leqslant j\leqslant n},$ as in the question, but it requires knowledge of linear algebra (of which my own knowledge is very patchy!): $\operatorname{det}(A^TA) \geqslant 0,$ with equality if and only if the columns of $A$ are linearly dependent. See, e.g., Gram determinant - Encyclopedia of Mathematics; also Theorem 8 of Hardy, Littlewood and Polya, Inequalities (1934, 1952). $\endgroup$ Jul 4, 2021 at 17:24

4 Answers 4


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.)

  • $\begingroup$ To illustrate, when $k = 2$ and $n = 3$: \begin{multline*} (a^2 + b^2)(c^2 + d^2) (e^2 + f^2) \geqslant (a^2 + b^2)(c^2e^2 + d^2f^2) = \\ (a^2 + b^2)((ce)^2 + (df)^2) \geqslant (a(ce) + b(df))^2 = (ace + bdf)^2. \end{multline*} $\endgroup$ Jul 3, 2021 at 19:01
  • $\begingroup$ @V.S.e.H. It's just using distributivity, and then dropping all the resulting product terms (which are all $\geqslant 0$) except those whose factors all have the same value of $i.$ In the example in the comment above: $$(c^2 + d^2) (e^2 + f^2) = c^2e^2 + c^2f^2 +d^2e^2 + d^2f^2 \geqslant c^2e^2 + d^2f^2.$$ It's as simple as that - unless I've goofed in some really daft way, of course! $\endgroup$ Jul 3, 2021 at 23:38
  • $\begingroup$ Yes, thanks, figured it out:) +1 Why do you think that the solution from Steele is complicated? I think it's pretty straightforward, and it works very well when generalizing Holder's inequality in a similar manner. $\endgroup$
    – V.S.e.H.
    Jul 4, 2021 at 11:24
  • 1
    $\begingroup$ I think Steele intentionally did that to keep in line, and help practice the key technique presented in the chapter, which is that of normalization. It is indeed a nice "sledge hammer" technique for a lot of inequalities. Nevertheless, the book is indeed filled with many weird, and overly complicated solutions to some problems (and with a lot of errors, some of which are not on the official errata). $\endgroup$
    – V.S.e.H.
    Jul 4, 2021 at 16:34
  • 1
    $\begingroup$ @V.S.e.H. Thanks - I shall keep that in mind when I go back to the book. (I didn't read very much of it.) $\endgroup$ Jul 4, 2021 at 16:44

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.


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .