= Solution
The <Harnack inequality on the unit disk> states that a nonnegative <harmonic function> satisfies
$$
\frac{1-|z|}{1+|z|}u(0)\le u(z)\le\frac{1+|z|}{1-|z|}u(0).
$$
One proof applies the <Poisson integral on the unit disk> on a circle of radius $R>|z|$: the <Poisson kernel on the circle> lies between $(R-|z|)/(R+|z|)$ and its reciprocal. The <mean value property for harmonic functions> identifies the mean boundary value with $u(0)$; let $R\uparrow1$.
A <harmonic conjugate> on the <simply connected> <unit disc> makes $H=u+iv$ <holomorphic>. The <Carathéodory derivative bound for the right half-plane> gives $|H'(0)|\le2u(0)$. The <Cauchy-Riemann equations> give $|H'|=|\nabla u|$, so $|\nabla u(0)|\le2u(0)$. For a fixed $z$, use the <disk automorphism>
$$
\psi(\zeta)=\frac{z+\zeta}{1+\overline z\zeta},\qquad \psi(0)=z,\qquad |\psi'(0)|=1-|z|^2.
$$
The <chain rule> and <conformality> multiply the <gradient> length by $|\psi'(0)|$. Applying the estimate at zero to $u\circ\psi$ proves the <sharp gradient bound for positive harmonic functions>:
$$
\boxed{|\nabla u(z)|\le\frac{2}{1-|z|^2}u(z),\qquad c_{\rm best}(z)=\frac{2}{1-|z|^2}.}
$$
Sharpness follows from $u(\zeta)=\operatorname{Re}[(1+\zeta)/(1-\zeta)]$, which has equality at zero, and then composing with $\psi^{-1}$. These extremizers are positive multiples of the <Poisson kernel on the circle> with its boundary pole fixed at one point.
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