From Intro to Calculus & Analysis I by R. Courant and Fritz John, pg. 504:
Problem: To solve the equation x=f(x), show how best to choose the content a such that the iteration $$x_{k+1}=x_k+a(x_k-f(x_k))$$ converges as rapidly as possible. (EDIT: In The Neighborhood of the Solution i.e. $\xi$)
My attempt (after some discussion with my friends):
Let $\phi(x_k)=x_{k+1}$, then a solution $\xi$ to $f(x)=x$ is also a solution to $\phi(x)=x$.
By MVT, $|x_{k+1}-\xi|=|x_k-\xi||\phi'(c_k)|=|x_k-\xi||a||1+\dfrac1a-f'(c_k)|$, where $c_k$ is between $x_k$ and $\xi$. Hence, $|x_{k+1}-\xi|=|x_1-\xi||a|^k\prod _{i=1}^k|1+\dfrac1a-f'(c_i)|$
From this point on, any attempt to progress end up being futile, almost always leading me back to where I started or giving me some really trivial result such as the absolute value of some quantity is not less than 0.
EDIT: Further Attempts:
$\forall 1\le i\le k, |1-f'(c_i)+1/a|\le |1+1/a-f'(\xi)|+|f'(\xi)-f'(c_i)|$, and if $f(x)\in C^2$ in the neighborhood of $\xi$, then there are some $\eta_1, ... , \eta_k$ such that $|f'(\xi)-f'(c_i)|=|\xi -c_i||f''(\eta_i)|$. Now since all $x_i$ is in the neighborhood of $\xi, |x_i-\xi|\lt \delta$ and also $|c_i-\xi|\le \delta$ for some fixed $\delta \gt 0$. Now, for the iteration to converge, we must have $|f'(\xi)|\le 1$ and therefore all the $|f'(c_i)|\lt 1$. Then $\dfrac{|f(c_i)-f(\xi)|}{\delta} \lt \dfrac{|f(x_i)-f(\xi)|}{|x_i-\xi|} \lt 1$