Edit: oop... misread... revised: The given inequality and Cauchy's formula for the second derivative $f''$, letting the large circle go to infinity, show that $f''(z)=0$, so $f$ is (not constant, but) linear. This is just a little extension of the argument for Liouville's theorem, so not really so much about maximum modulus, perhaps.
Edit-edit: explicitly, by the Cauchy integral formula for the derivatives, $f''(z)={2!\over 2\pi i}\int_\gamma {f(\zeta)\,d\zeta\over (\zeta-z)^3}$, where $\gamma$ is a large circle of radius $R$. The numerator is bounded by $R^{3/2}$, and the denominator is essentially $R^3$. The length of the curve is $2\pi R$, so the integral expressing the second derivative is bounded by a constant multiple of $1/R^{1/2}$, which goes to $0$ as $R$ goes to $+\infty$.