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For a holomorphic function $h:\mathbb{D}\to \mathbb{D}$, the Schwarz-Pick lemma statesEquality in the first inequality for distinct points, or in the derivative inequality at one point, occurs exactly when $h$ is an automorphism of the unit disk.

Here is a proof, including the underlying Schwarz lemma. If $u:\mathbb{D}\to \mathbb{D}$ is holomorphic and $u(0)=0$, then $u(z)/z$ has a removable singularity at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle on this circle, followed by $r\uparrow 1$, gives $|u(z)|\le |z|$ and $|u^{\prime }(0)|\le 1$. Equality at a nonzero point, or $|u^{\prime }(0)|=1$, forces $u(z)/z$ to be a constant of modulus one by the same maximum modulus principle.

Set ${\varphi }_a(z)=(z-a)/(1-\overline{a}z)$, an automorphism of the unit disk. The composite ${\varphi }_{h(a)}\circ h\circ {\varphi }_a^{-1}$ fixes zero. The Schwarz lemma gives the first inequality, while its derivative at zero gives the second. The equality conclusion follows because a composite attaining equality is a rotation. Conversely, every automorphism of the unit disk attains equality. This also proves contraction of the hyperbolic metric.

The boundary lower bound first guarantees a zero inside the unit circle. If $f(0)=0$, there is nothing to prove about existence. Otherwise, if $f$ had no zeros on the closed unit disc, $1/f$ would be holomorphic on a neighborhood of it and satisfy $|1/f|\le 1/A$ on the boundary. The maximum modulus principle would give $|f(0)|\ge A$, contradicting the hypothesis. The boundary lower bound excludes boundary zeros, so there is a zero $a$ with $|a|<1$.

The upper boundary bound and the maximum modulus principle give $|f|\le B$ on the unit circle and its interior. The function is nonconstant under the stated hypotheses, so $f/B$ takes its interior values in the open unit disc. Apply the Schwarz-Pick lemma to the points $a$ and $0$:This is the zero location bound from Schwarz-Pick, and it applies to every zero inside the unit circle.

The bound is sharp. Given $0<a<1$ and $B>0$, chooseThis holomorphic function has no pole in ${\mathbb{D}}_R$, has constant modulus $B$ on the unit circle, and satisfies $|f(0)|=Ba<B$. With $A=B$, its zero at $a$ gives equality. The case $a=0$ is also attained by $f(z)=Bz$.