# Cauchy -Schwarz for double summation

In $R^n$ with the standard inner product, the Cauchy-Schwarz inequality is $$\left ( \sum_{i=1}^n a_i b_i \right)^2 \le \left ( \sum_{i=1}^n a_i ^2\right) \left ( \sum_{i=1}^n b_i^2 \right) .$$

Is there an analog of Cauchy-Schwarz inequality for double summation? Say, how does one apply Cauchy-Schwarz to something like: $$\sum_{i=1}^n\sum_{j=1}^n a_i b_j ~?$$

• @Bungo You are right. fixed it. Nov 19, 2017 at 22:40

Rewrite the sum you are curious about as $$\Big( \sum_{i=1}^n a_i \Big) \Big( \sum_{j=1}^n b_j \Big)$$ Then, I'm not really sure what you want to use Cauchy-Schwarz for here, but you could say $$\Big( \sum_{i=1}^n a_i \Big)^2 \Big( \sum_{j=1}^n b_j \Big)^2 \leq n^2 \Big( \sum_{i=1}^n a_i^2 \Big) \Big( \sum_{j=1}^n b_j^2 \Big)$$

• well, can I avoid using Cauchy-Schwarz? how do I find an upper bound for the sum I'm curious about. Nov 19, 2017 at 22:42
• You can bound it in many different ways - as stated, the form is too abstract to give one perfect bound for it, though. We would need more info. Nov 19, 2017 at 22:44
• You don't need Cauchy-Schwarz for this, @L.bronze. Just note that $$\left|\sum a_i\right| \le \sum |a_i| \le \sum \|a\|=n\|a\|,$$ where $\|a\|^2 = \sum a_i^2$. Oct 8, 2019 at 16:37

There is the following inequality, which we can prove by C-S. $$\sum_{i=1}^n a_i b_i+\sqrt{\sum_{i=1}^n a_i^2\sum_{i=1}^n b_i^2}\ge\ \frac{2}{n}\sum_{i=1}^n\sum_{j=1}^n a_ib_j$$

• Can you show your proof to this inequality? I have a method by BW, but it is so complicated that I don't know if it is valid. May 1 at 4:26
• There is a proof here, artofproblemsolving.com/community/c6h2506098p21184363 but I can't understand it. May 1 at 4:41
• @youthdoo Post your question in this forum. I have a proof in one line. May 1 at 4:53
• In MSE or AoPS do you mean? May 1 at 4:56
• @youthdoo I meant MSE May 1 at 5:03