Best constant in an integral inequality Which is the smallest constant $B_d$ such that the following inequality 
 $$\left|\int_{0}^{1}t^d(1-t)\psi(t) dt\right|^2\le
 B_d\int_{0}^{1}t^d|\psi(t)|^2dt$$ holds, provided that $\psi(t)$ is a polynomial?

A hint: Consider the scalar product
$$\langle u,v\rangle:=\int_0^1 u(t) v(t)\ t^d\ dt\ .$$
