Prove $SST=SSE+SSR$ Prove $$SST=SSE+SSR$$
I start with $$SST= \Sigma (y_i-\bar{y})^2=...=SSE+SSR+ \Sigma 2( y_i-y_i^*)(y_i^*-\bar{y} )$$ and I don't know how to prove that  $\Sigma 2( y_i-y_i^*)(y_i^*-\bar{y} )=0$

a note on notation: the residuals $e_i$ is $e_i=y_i-y_i^*$. A more common notation is $\hat{y}$.
 A: When an intercept is included in linear regression(sum of residuals is zero), $SST=SSE+SSR$.
prove
$$
\begin{eqnarray*}
SST&=&\sum_{i=1}^n (y_i-\bar y)^2\\&=&\sum_{i=1}^n (y_i-\hat y_i+\hat y_i-\bar y)^2\\&=&\sum_{i=1}^n (y_i-\hat y_i)^2+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)+\sum_{i=1}^n(\hat y_i-\bar y)^2\\&=&SSE+SSR+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)
\end{eqnarray*}
$$
Just need to prove last part is equal to 0: 
$$\begin{eqnarray*}
\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)&=&\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)(\beta_0+\beta_1x_i-\bar y)\\&=&(\beta_0-\bar y)\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)+\beta_1\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)x_i
\end{eqnarray*}
$$
In Least squares regression, the sum of the squares of the errors is minimized.
$$
SSE=\displaystyle\sum\limits_{i=1}^n \left(e_i \right)^2= \sum_{i=1}^n\left(y_i - \hat{y_i} \right)^2= \sum_{i=1}^n\left(y_i -\beta_0- \beta_1x_i\right)^2
$$
Take the partial derivative of SSE with respect to $\beta_0$ and setting it to zero.
$$
\frac{\partial{SSE}}{\partial{\beta_0}} = \sum_{i=1}^n 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 (-1) = 0
$$
So
$$
\sum_{i=1}^n \left(y_i - \beta_0 - \beta_1x_i\right)^1 (-1) = 0
$$
Take the partial derivative of SSE with respect to $\beta_1$ and setting it to zero.
$$
\frac{\partial{SSE}}{\partial{\beta_1}} = \sum_{i=1}^n 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 (-x_i) = 0
$$
So
$$
\sum_{i=1}^n \left(y_i - \beta_0 - \beta_1x_i\right)^1 x_i = 0
$$
Hence,
$$
\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=(\beta_0-\bar y)\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)+\beta_1\sum_{i=1}^n(y_i-\beta_0-\beta_1x_i)x_i=0
$$
$$SST=SSE+SSR+2\sum_{i=1}^n(y_i-\hat y_i)(\hat y_i-\bar y)=SSE+SSR$$
similar question: 
https://stats.stackexchange.com/a/401299/243636
A: The principle underlying least squares regression is that the sum of the squares of the  errors is minimized. We can use calculus to find equations for the parameters $\beta_0$ and $\beta_1$ that minimize the sum of the squared errors.
Let $S = \displaystyle\sum\limits_{i=1}^n \left(e_i \right)^2= \sum \left(y_i - \hat{y_i} \right)^2= \sum \left(y_i - \beta_0 - \beta_1x_i\right)^2$ 
We want to find $\beta_0$ and $\beta_1$ that minimize the sum, $S$. We start by taking the partial derivative of $S$ with respect to $\beta_0$ and setting it to zero.
$$\frac{\partial{S}}{\partial{\beta_0}} = \sum 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 (-1) = 0$$
notice that this says, 
$$\begin{align}\sum \left(y_i - \beta_0 - \beta_1x_i\right) &= 0 \\
       \sum \left(y_i - \hat{y_i} \right) &= 0 \qquad (eqn. 1)\end{align}$$
Hence, the sum of the residuals is zero (as expected). Rearranging and solving for $\beta_0$ we arrive at,
$$\sum \beta_0 = \sum y_i -\beta_1 \sum x_i $$
$$n\beta_0 = \sum y_i -\beta_1 \sum x_i $$
$$\beta_0 = \frac{1}{n}\sum y_i -\beta_1 \frac{1}{n}\sum x_i $$
now taking the partial of $S$ with respect to $\beta_1$ and setting it to zero we have, 
$$\frac{\partial{S}}{\partial{\beta_1}} = \sum 2\left(y_i - \beta_0 - \beta_1x_i\right)^1 (-x_i) = 0$$ 
and dividing through by -2 and rearranging we have,
$$\sum x_i \left(y_i - \beta_0 - \beta_1x_i\right) = 0$$
$$\sum x_i \left(y_i - \hat{y_i} \right) = 0$$
but, again we know that $\hat{y_i} = \beta_0 + \beta_1x_i$. Thus, $x_i = \frac{1}{\beta_1}\left( \hat{y_i} - \beta_0 \right) = \frac{1}{\beta_1}\hat{y_i} -\frac{\beta_0}{\beta_1}$. Substituting this into the equation above gives the desired result. 
$$\sum x_i \left(y_i - \hat{y_i} \right) = 0 $$
$$\sum \left(\frac{1}{\beta_1}\hat{y_i} - \frac{\beta_0}{\beta_1}\right) \left(y_i - \hat{y_i} \right) = 0$$
$$\frac{1}{\beta_1}\sum \hat{y_i} \left(y_i - \hat{y_i} \right) - \frac{\beta_0}{\beta_1} \sum \left(y_i - \hat{y_i} \right)= 0$$
Now, the second term is zero (by eqn. 1) and so, we arrive immediately at the desired result: 
$$\sum \hat{y_i} \left(y_i - \hat{y_i} \right) = 0  \qquad (eqn. 2)$$
Now, let's use eqn. 1 and eqn. 2 to show that 
$\sum \left(\hat{y_i} - \bar{y_i} \right) \left( y_i - \hat{y_i} \right) = 0$ - which was your original question. 
$$\sum \left(\hat{y_i} - \bar{y_i} \right) \left( y_i - \hat{y_i} \right) = \sum \hat{y_i} \left( y_i - \hat{y_i} \right)  - \bar{y_i} \sum \left( y_i - \hat{y_i} \right) = 0$$
A: If you have already found the formulas for $b_0$ and $b_1$, but you are having trouble proving that $\sum_{i=1}^n (y_i - \hat{y}_i)(\hat{y}_i - \bar{y}) = 0$, I think that the following proof is an interesting one:
\begin{aligned}
\sum_{i=1}^n (y_i - \hat{y}_i)(\hat{y}_i - \bar{y}) &= \sum_{i=1}^{n}(y_i - \bar{y} -b_1 (x_i - \bar{x}))(\bar{y} + b_1 (x_i - \bar{x})-\bar{y}) \\
&= b_1 \sum_{i=1}^{n} (y_i - \bar{y})(x_i - \bar{x}) - b_1^2\sum_{i=1}^{n}(x_i - \bar{x})^2 \\
&= b_1 \frac{\sum_{i=1}^{n}(y_i -\bar{y})(x_i - \bar{x})}{\sum_{i=1}^{n}(x_i - \bar{x})^2} \sum_{i=1}^{n}(x_i - \bar{x})^2 - b_1^2\sum_{i=1}^{n}(x_i - \bar{x})^2 \\
&= b_1^2 \sum_{i=1}^{n}(x_i - \bar{x})^2 - b_1^2 \sum_{i=1}^{n}(x_i - \bar{x})^2 \\
&= 0
\end{aligned}
A: $$2\sum(y_i-y_i^*)(y_i^*-\bar{y})$$
$$= 2\sum[y_i(y_i^*-\bar{y})-y_i^*(y_i^*-\bar{y})]$$
$$= 2\sum Ye_i - 2\bar{Y}\sum e_i$$
$$= 0$$
