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).
Best Answer
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.)