From the mean value equality of harmonic functions, if $\Delta u=0$, $$u(0)={|\partial B(0,1)|}^{-1} \int_{\partial B(0,1)} u(y)dS_y={|B(0,1)|^{-1}} \int_{B(0,1)} u(y)dy.$$ So $$\int_{\partial B(0,1)} u(y)dS_y=|\partial B(0,1)|u(0)=n\omega_nu(0)$$ while $$\int_{B(0,1)} u(y)dy=\omega_nu(0).$$ So whenever $n\ge 2$, the integral on the surface is bigger than the one in the ball.
My question is, does that hold for subharmonic functions? If $ \ \Delta v\ge 0 \ $ (for example, $v=u^p$, $p>1$), which integral is bigger?
Update: I found a result from a lecture note without proof. It seems that my question is not strong enough. It writes, $${|\partial B(0,1)|}^{-1} \int_{\partial B(0,1)} u^p(y)dS_y\ge {|B(0,1)|^{-1}} \int_{B(0,1)} u^p(y)dy.$$ But I have no idea of how to prove this one.
Thank you for your help!