Integral inequality - lower bound on $L^1$ norm.

I was wondering if one can make an estimate of form:

Assume $f\in C^\infty(\overline{\Omega})$ where $\Omega$ is a bounded domain in $\mathbb{}R^d$. Is there a constant $C>0$ independent of $f$ such that: $$\int\limits_{B_r(x)}\int\limits_0^1(1-t)^{d-1}|f(y+t(x-y))|\ \mathrm{dt}\mathrm{dy}\leq C||f||_{L^1(\Omega)},$$ where $B_r(x)$ is a ball in $\Omega$?

If there were factor $(1-t)^d$ instead of $(1-t)^{d-1}$, the estimate would follow immediately from substition and Fubini theorem. However, that's not the case I need.

Is it possible at all? Thanks for any advice.

• Yes, I've just edited my post - sorry for that:) – user1321324 Nov 8 '15 at 12:25
• Try: Applying the Chebyshev second inequality or sometimes known as Gruss inequality. – mwomath Jun 8 '17 at 12:31