Let $h$ be a positive harmonic function on $\Delta (0,\rho )=\lbrace z\in\mathbb{C} : |z|\leq \rho \rbrace$. I wish to show that $|\nabla h(z)|\leq \frac{2\rho}{\rho ^2-|z|^2}h(z)$.
Since $h$ is harmonic on a domain, we know that there exist an holomorphic function $f$ such that $h=\mathfrak {R} (f)$. It is maybe easier to work with an holomorphic function in order to establish the inequality. We have $f'(z)=h_x-ih_y$, so $|\nabla h(z)|=|f'(z)|$. I tried to obtain an upper bound by using Cauchy formula but I did'nt succeed. It is certainly not a good way since the upper bound depends on $|f(z)|$ and not $h(z)$.
Begin with the special case: $h$ is positive and harmonic on $\Delta(0,1)$, we want $|\nabla h(0)|\le 2 h(0)$. The Poisson formula is $$h(z) = \frac{1}{2\pi} \int_{|\zeta|=1} \frac{1-|z|^2}{|z-\zeta|^2} h(\zeta)\,|d\zeta| \tag1$$ Observe that as $z\to 0$, $$\frac{1-|z|^2}{|z-\zeta|^2}\le \frac{1}{(1- |z|)^2} = 1+ 2 |z| +o(|z|) \tag2$$ Use (2) in (1) to obtain $$h(z) \le (1+2|z|) h(0) + o(|z|)$$ which gives the desired result $|\nabla h(0)|\le 2 h(0)$.
To get the general form, consider $h\circ \phi$ where $$\phi(w) = \rho\frac{w+z/\rho}{1+w \bar z /\rho}$$ is a holomorphic map of $\Delta(0,1)$ onto $\Delta(0,\rho)$ sending $0$ to $z$.