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?

Presumably $d > -1$ so the integrals converge.
Consider the Hilbert space $L^2(\mu)$ where $\mu$ is the measure $\mu(dt) = t^d \ dt$ on $[0,1]$.  Your inequality says $|\langle 1-t, \psi\rangle|^2 \le B_d \|\psi\|^2$.  The Cauchy-Schwarz inequality says the inequality is true for all $\psi \in L^2(\mu)$, in particular for polynomials, with $$B_d = \|1-t\|^2 = \int_0^1 t^d (1-t)^2\ dt = \dfrac{2}{(d+1)(d+2)(d+3)}$$
and this is best possible (since you could take $\psi(t) = 1-t$).
