mattmns
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Here is the question from the book:
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Let n\geq1 and let a_1,...,a_n and b_1,...,b_n be real numbers. Verify the identity:
\left(\sum_{i=1}^n{a_ib_i}\right)^2 + \frac{1}{2}\sum_{i=1}^n{\sum_{j=1}^n{\left(a_ib_j-a_jb_i\right)^2}} = \left(\sum_{i=1}^n{a_i^2}\right)\left(\sum_{j=1}^n{b_j^2}\right)
and conclude the Cauchy-Schwartz inequality:
\left|\sum_{i=1}^n{a_ib_i}\right| \leq \left(\sum_{i=1}^n{a_i^2}\right)^{1/2} \left(\sum_{j=1}^n{b_j^2}\right)^{1/2}
Then use Cauchy-Schwartz to prove the triangle inequality:
\left(\sum_{i=1}^n{(a_i^2+b_i^2)}\right)^{1/2} \leq \left(\sum_{i=1}^n{a_i^2}\right)^{1/2} + \left(\sum_{j=1}^n{b_j^2}\right)^{1/2}
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I have been trying to mess around with the first one, and see what is going on, but it is looking extremely ugly, even with n=3.
Should I be trying to prove these by induction? Or are there some ways to manipulate these things easily? I guess the problem is almost notation, or that I don't know how to manipulate sums and squares properly. Any ideas would be appreciated. Thanks!
------
Let n\geq1 and let a_1,...,a_n and b_1,...,b_n be real numbers. Verify the identity:
\left(\sum_{i=1}^n{a_ib_i}\right)^2 + \frac{1}{2}\sum_{i=1}^n{\sum_{j=1}^n{\left(a_ib_j-a_jb_i\right)^2}} = \left(\sum_{i=1}^n{a_i^2}\right)\left(\sum_{j=1}^n{b_j^2}\right)
and conclude the Cauchy-Schwartz inequality:
\left|\sum_{i=1}^n{a_ib_i}\right| \leq \left(\sum_{i=1}^n{a_i^2}\right)^{1/2} \left(\sum_{j=1}^n{b_j^2}\right)^{1/2}
Then use Cauchy-Schwartz to prove the triangle inequality:
\left(\sum_{i=1}^n{(a_i^2+b_i^2)}\right)^{1/2} \leq \left(\sum_{i=1}^n{a_i^2}\right)^{1/2} + \left(\sum_{j=1}^n{b_j^2}\right)^{1/2}
----------------
I have been trying to mess around with the first one, and see what is going on, but it is looking extremely ugly, even with n=3.
Should I be trying to prove these by induction? Or are there some ways to manipulate these things easily? I guess the problem is almost notation, or that I don't know how to manipulate sums and squares properly. Any ideas would be appreciated. Thanks!