It's a lot easier than that if you have theorems for manipulating summations.
\(\displaystyle \displaystyle \sum(x_i - \bar x)(y_i - \bar y) = \sum\left( (x_i - \bar x)y_i + (x_i - \bar x)\bar y \right) = \sum (x_i - \bar x)y_i + \sum (x_i - \bar x)\bar y =\)
\(\displaystyle \displaystyle \sum (x_i - \bar x)y_i + \bar y \sum(x_i - \bar x) = \sum (x_i - \bar x)y_i + 0 = \sum(x_i - \bar x)y_i \)
Yes. I thought about it some more and came back to edit and found your response. I cannot imagine now why I thought induction would be necessary because, as shown below, my approach generalizes. Of course, your approach is more elegant and avoids induction as well.
\(\displaystyle \bar x = \dfrac{\displaystyle \sum_{j=1}^n x_j}{n} \implies n \bar x = \displaystyle \sum_{j=1}^n x_j.\)
\(\displaystyle \displaystyle \therefore \left ( \sum_{j=1}^n (x_j - \bar x)(y_j - \bar y) \right ) =\)
\(\displaystyle \displaystyle \left ( \sum_{j=1}^n x_jy_j - x_j \bar y - \bar x y_j + \bar x \bar y) \right ) = \left ( \sum_{j=1}^n x_jy_j - \bar x y_j - x_j \bar y + \bar x \bar y) \right ) =\)
\(\displaystyle \displaystyle \left ( \sum_{j=1}^n x_jy_j - \bar x y_j \right ) - \left ( \sum_{j=1}^n x_ j \bar y \right ) + \left ( \sum_{j=1}^n \bar x \bar y) \right ) =\)
\(\displaystyle \displaystyle \left ( \sum_{j=1}^n x_jy_j - \bar x y_j \right ) - \left \{ \bar y * \left ( \sum_{j=1}^n x_ j \right ) \right \} + n \bar x \bar y =\)
\(\displaystyle \displaystyle \left ( \sum_{j=1}^n x_jy_j - \bar x y_j \right ) - \{ \bar y * (n \bar x)\} + n \bar x \bar y =\)
\(\displaystyle \displaystyle \left ( \sum_{j=1}^n x_jy_j - \bar x y_j \right ) - n \bar x \bar y + n \bar x \bar y =\)
\(\displaystyle \displaystyle \left ( \sum_{j=1}^n x_jy_j - \bar x y_j \right ) =\)
\(\displaystyle \displaystyle \left ( \sum_{j=1}^n (x_j - \bar x )y_j \right ).\)
No induction needed. The n = 2 case generalizes directly, and the only theorem on sums needed is that relating to the sum of a constant.