1
$\begingroup$

Suppose I have the following inequality:

$$\frac{N_1^2S_1^2 + N_2^2S_2^2 + \cdots + N_n^2S_n^2}{(N_1 + N_2 + \cdots + N_n)^2} \leq \min(S_1^2, S_2^2, \ldots, S_n^2)$$

Where:

  • $N_1, N_2, \ldots, N_n \in \mathbb{N}$
  • $S_1^2, S_2^2, \ldots, S_n^2 \in \mathbb{R}^+$

I am trying to prove whether this inequality is True or False for the general case

I am not sure how to prove this for the general case - but I think I was able to show these for a specific case where $S_1 = S_2 = \cdots = S_n = S$.

  • Suppose we define $W_i = \frac{N_i}{\sum_{j=1}^n N_j}$

  • It then follows that $\sum_{i=1}^n W_i = 1$ and $\sum_{i=1}^n W_i^2 \leq 1$

  • This means we can re-write the inequality as $\sum_{i=1}^n W_i^2S_i^2 \leq \min(S_1^2, S_2^2, \ldots, S_n^2)$

  • Since $S_1 = S_2 = \cdots = S_n = S$ , then $\sum_{i=1}^n W_i^2S_i^2 = \sum_{i=1}^n W_i^2S^2 $

  • This means that we can again re-write the inequality as $\sum_{i=1}^n W_i^2S^2 \leq \min(S^2) = S^2$

  • Re-writing the above step, we have: $S^2\sum_{i=1}^n W_i^2 \leq S^2$ and $\sum_{i=1}^n W_i^2 \leq 1$

  • As a result, I think the inequality has been proven for a specific case when $S_1 = S_2 = \cdots = S_n = S$.

My Question: However, I am still interested in learning if this inequality can be proven (yes or no) for the general case where $S_1^2 \neq S_2^2 \neq \cdots \neq S_n^2$

Can someone please show me how to do this? Is the Cauchy-Schwartz inequality needed for this?

Thanks!

$\endgroup$
3
  • 3
    $\begingroup$ This cannot be true in the general case, because with increasing (e.g.) $S_1$ the left-hand side becomes arbitrarily large, whereas the right-hand side does not change. $\endgroup$
    – Martin R
    Commented May 22, 2023 at 7:17
  • $\begingroup$ If all $S_i$ are equal then the inequality is $N_1^2 + \cdots + N_n^2\le (N_1 + \cdots + N_n)^2$, which is true and follows (e.g.) from expanding the right-hand side. $\endgroup$
    – Martin R
    Commented May 22, 2023 at 7:23
  • $\begingroup$ The best bound you can get for the left side (in general) is $\max \{S_1^{2},S_2^{2},...,S_n^{2}\}$. $\endgroup$ Commented May 22, 2023 at 7:40

2 Answers 2

2
$\begingroup$

The inequality cannot be true in the general case, because with increasing (e.g.) $S_1$ the left-hand side becomes arbitrarily large, whereas the right-hand side does not change.

If all $S_j = S$ are equal then

$$ \frac{N_1^2 S_1^2 + \cdots + N_n^2 S_n^2}{(N_1 + \cdots + N_n)^2} = \frac{N_1^2 + \cdots + N_n^2 }{(N_1 + \cdots + N_n)^2} \cdot S^2 \le S^2 $$ because $$ N_1^2 + \cdots + N_n^2 \le (N_1 + \cdots + N_n)^2 $$ as can be seen by expanding the right-hand side.

An upper bound for the general case is obtained in a similar way: $$ \frac{N_1^2 S_1^2 + \cdots + N_n^2 S_n^2}{(N_1 + \cdots + N_n)^2} \le \frac{N_1^2 + \cdots + N_n^2 }{(N_1 + \cdots + N_n)^2} \cdot \max(S_1^2, \ldots, S_n^2) \le \max(S_1^2, \ldots, S_n^2) \, . $$

$\endgroup$
2
  • $\begingroup$ Congratulations on 100k, I believe that’s recent $\endgroup$
    – FShrike
    Commented May 24, 2023 at 8:50
  • 1
    $\begingroup$ @FShrike: Thank you. $\endgroup$
    – Martin R
    Commented May 24, 2023 at 8:53
1
$\begingroup$

The inequality in question does not generally hold. However, the following does$$\max_{\{N_i\}_{i=1}^N}\frac{N_1^2S_1^2 + N_2^2S_2^2 + \cdots + N_n^2S_n^2}{(N_1 + N_2 + \cdots + N_n)^2}\le\max\{S_1^2,\cdots ,S_n^2\},$$which we will prove.

WLOG, we assume $S_1^2\le S_2^2\le\cdots \le S_n^2$. We try to maximize the LHS of the inequality w.r.t. $N_i$s. Hence, $$ \max_{\{N_i\}_{i=1}^N}\frac{N_1^2S_1^2 + N_2^2S_2^2 + \cdots + N_n^2S_n^2}{(N_1 + N_2 + \cdots + N_N)^2} { = \max_{M\ge n}\max_{\{N_i\}_{i=1}^n,\sum_{i=1}^nN_i=M}\frac{N_1^2S_1^2 + N_2^2S_2^2 + \cdots + N_n^2S_n^2}{(N_1 + N_2 + \cdots + N_n)^2} \\= \max_{M\ge n}\max_{\{N_i\}_{i=1}^n,\sum_{i=1}^nN_i=M}\frac{N_1^2S_1^2 + N_2^2S_2^2 + \cdots + N_n^2S_n^2}{M^2} \\= \max_{M\ge n}\frac{(M-n+1)^2S_n^2+\sum_{i=1}^{n-1}S_i^2}{M^2} \\= S_n^2\cdot\max_{M\ge n}\frac{(M-n+1)^2+\sum_{i=1}^{n-1}\frac{S_i^2}{S_n^2}}{M^2}. } $$ The function $f(x)=\frac{(x-a)^2+b}{x^2}$ has a minimum over $(0,\infty)$ at $x=a+b/a$. Applying this fact to $f(M)=\frac{(M-n+1)^2+\sum_{i=1}^{n-1}\frac{S_i^2}{S_n^2}}{M^2}$, we have $a=n-1$ and $b=\sum_{i=1}^{n-1}\frac{S_i^2}{S_n^2}$ and the minimum resides in $$ M^*=n-1+\sum_{i=1}^{n-1}\frac{S_i^2}{(n-1)S_n^2}. $$ It is easy to observe that $n-1\le M^*\le n$. Therefore, $\frac{(M-n+1)^2+\sum_{i=1}^{n-1}\frac{S_i^2}{S_n^2}}{M^2}$ is strictly increasing for $M\ge n$. Therefore, $$ \max_{M\ge n}\frac{(M-n+1)^2+\sum_{i=1}^{n-1}\frac{S_i^2}{S_n^2}}{M^2}=\lim_{M\to\infty} \frac{(M-n+1)^2+\sum_{i=1}^{n-1}\frac{S_i^2}{S_n^2}}{M^2}=1. $$

We finally conclude $$ \max_{\{N_i\}_{i=1}^N}\frac{N_1^2S_1^2 + N_2^2S_2^2 + \cdots + N_n^2S_n^2}{(N_1 + N_2 + \cdots + N_n)^2}\le\max\{S_1^2,\cdots ,S_n^2\}. $$

$\endgroup$
2
  • 1
    $\begingroup$ Should it be $(N_1 + N_2 + \cdots + N_n)^2$ in the denominator on the left? In that case a simple proof would be $$ \frac{N_1^2S_1^2 + N_2^2S_2^2 + \cdots + N_n^2S_n^2}{(N_1 + N_2 + \cdots + N_n)^2} \leq \frac{N_1^2 + N_2^2 + \cdots + N_n^2}{(N_1 + N_2 + \cdots + N_n)^2} \cdot \max S_i^2 \le \max S_i^2 $$ since $N_1^2 + \cdots + N_n^2\le (N_1 + \cdots + N_n)^2$. $\endgroup$
    – Martin R
    Commented May 22, 2023 at 9:28
  • 1
    $\begingroup$ Yes. This proof for the modified inequality is remarkably simpler. $\endgroup$ Commented May 22, 2023 at 9:38

You must log in to answer this question.

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