= Solution
For a <holomorphic function> $h:\mathbb D\to\mathbb D$, the <Schwarz-Pick lemma> states
$$
\left|\frac{h(z)-h(a)}{1-\overline{h(a)}h(z)}\right|
\leq\left|\frac{z-a}{1-\overline a z}\right|,
\qquad
\frac{|h'(a)|}{1-|h(a)|^2}\leq\frac1{1-|a|^2}.
$$
Equality 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 $r\uparrow1$, gives $|u(z)|\leq|z|$ and $|u'(0)|\leq1$. Equality at a nonzero point, or $|u'(0)|=1$, forces $u(z)/z$ to be a constant of modulus one by the same <maximum modulus principle>.
Set $\phi_a(z)=(z-a)/(1-\overline a z)$, an <automorphism of the unit disk>. The composite $\phi_{h(a)}\circ h\circ\phi_a^{-1}$ 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 $|1/f|\leq1/A$ on the boundary. The <maximum modulus principle> would give $|f(0)|\geq A$, 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 $|f|\leq B$ 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 $0$:
$$
\boxed{|a|\geq\frac{|f(0)|}{B}.}
$$
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 $B>0$, choose
$$
f(z)=B\frac{z-a}{1-az},\qquad 1<R<1/a.
$$
This <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 $f(z)=Bz$.
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