I have to prove that $$\sum_{k=0}^{n}\left\lvert x-\frac{k}{n}\right\rvert\binom{n}{k}x^k(1-x)^{n-k} \le \frac{1}{2\sqrt{n}} $$ where $n\in \mathbb{N}$ and $x \in [0,1]$
I have already proven that $\displaystyle \sum_{k=0}^{n}\left(x-\frac{k}{n}\right)^2\binom{n}{k}x^k(1-x)^{n-k} \le \frac{1}{4n}$ and that I have to use the Cauchy Schwartz formula but I'm stuck because it never yields the wanted formula.
Thank you.
In the language of random variables, you already have $E |x - Y/n|^2 \le \frac{1}{4n}$, where $Y$ is a binomial random variable of parameters $x$ and $n$. Using that $E Z^2 \geq (E Z)^2$ leads to the desired inequality. This inequality is indeed a consequence of Cauchy-Schwarz, since it can be obtained by: $$ EZ = E[1 \cdot Z] \le E[1^2]^{1/2} \cdot E[Z^2]^{1/2}.$$