In foundations of cryptography the author is trying to demonstrate the success an adversarial algorithm A1 will have using just random guessing. A1 will, given the output of some function f will try and produce its pre-image by guessing, and the author shows this probability to be greater than 2^-n. Is there any reason for this? It also says that this sum is minimized when all the values to be squared are equal why is this obvious?
Yes, use the Cauchy-Schwarz inequality $(x \cdot y)^2\leq (x \cdot x)(y \cdot y)$ with equality if and only if $x$ and $y$ are parallel.
In particular, take $x = (x_1, \dots, x_n)$ and $y = (1, \dots, 1)$, so $$x \cdot y = \sum_{i=1}^n x_i = 1$$ $$x \cdot x = \sum_{i=1}^n x_i^2$$ $$y \cdot y = n$$ hence $$\sum_{i=1}^n x_i^2 \geq \frac{1}{n}.$$