Suppose that $f$ is continuous on $[0,1]$ and $$ \int_0^1 x^kf(x)\ dx=0$$ for $k=0,1,...,n-1$, $$\ \int_0^1x^nf(x) \ dx=1$$ Prove that there exists $\xi\in(0,1)$ such that $|f(\xi)|\geq 2^n(n+1)$.
So my idea is to use proof by contradiction, but nothing comes out