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