# Proving the General Case of this Quadratic Inequality?

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!

• 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. Commented May 22, 2023 at 7:17
• 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. Commented May 22, 2023 at 7:23
• The best bound you can get for the left side (in general) is $\max \{S_1^{2},S_2^{2},...,S_n^{2}\}$. Commented May 22, 2023 at 7:40

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: $$\frac{N_1^2 S_1^2 + \cdots + N_n^2 S_n^2}{(N_1 + \cdots + N_n)^2} \le \frac{N_1^2 + \cdots + N_n^2 }{(N_1 + \cdots + N_n)^2} \cdot \max(S_1^2, \ldots, S_n^2) \le \max(S_1^2, \ldots, S_n^2) \, .$$

• Congratulations on 100k, I believe that’s recent Commented May 24, 2023 at 8:50
• @FShrike: Thank you. Commented May 24, 2023 at 8:53

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

• 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 $N_1^2 + \cdots + N_n^2\le (N_1 + \cdots + N_n)^2$. Commented May 22, 2023 at 9:28
• Yes. This proof for the modified inequality is remarkably simpler. Commented May 22, 2023 at 9:38