Prove $\max_{|z|=1} |f(z) - e^z/z| \geq 1$ for all holomorphic function in $\bar{D}$ Problem statement: Prove $$\max_{|z|=1} |f(z) - e^z/z| \geq 1$$ for all function $f$ holomorphic in the closed unit disk $\bar{D}$.
I guess I can put it into $\max_{|z|=1} |zf(z) - e^z| \geq 1$, but what tool do I use then?
 A: What about Cauchy's integral formula? Let $g(z) = zf(z) - e^z$ so that
$$-1 = g(0) = \frac 1{2\pi i} \int_{|z|=1} \frac{g(z)}{z} \, dz.$$ Modulus and the triangle inequality give you
$$1 \le \frac 1{2\pi} \int_{|z|=1} |g(z)| \, dz.$$ Is is possible that $|g(z)| < 1$ for all $|z| = 1$?
A: An answer has already been endorsed, but I'd like to provide a different one for other people since questions of this type scream out Rouché's to me.
Suppose otherwise so that $|f(z)-\frac{e^z}{z}|<1$ on $|z|=1$. In particular, multiplying both sides by $|z|=1$ gives $|zf(z)-e^z|<1$, so Rouché's theorem says that $zf(z)-e^z+1$ and 1 have the same number of zeroes in $\mathbb{D}$, as both functions are analytic, hence have no poles. Then there is no $z_0\in\mathbb{D}$ with $z_0f(z_0)+1 = e^{z_0}$. However, we know that choosing $z_0=0$ satisfies this equation, as $0\cdot f(0)+1 = 1$ since $f$ is well-defined at $z=0$. Therefore, our original assumption that $|f(z)-\frac{e^z}{z}|<1$ for all $z\in\partial\mathbb{D}$ must be false, so there is some $z\in\partial\mathbb{D}$ with $|f(z)-\frac{e^z}{z}|\geq1$. Equivalently, we must have that $\max\limits_{|z|=1}|f(z)-\frac{e^z}{z}|\geq1$.
edit: I should also include that we know the two functions $zf(z)-e^z+1$ and 1 are nonzero on $\partial\mathbb{D}$ since $|zf(z)-e^z|<1$ forces $zf(z)-e^z+1$ to be strictly positive (the inf of the real part of this quantity would be achieved when $zf(z)-e^z=-1$, but the strict inequality rules this out). On the other hand, 1 is clearly nonzero on the boundary, so the assumptions of Rouché's theorem are all satisfied.
A: By the Rouché theorem, $|f(z)-e^z/z|<1=|1/z|$ for $|z|=1$ means that the balance of roots and poles (with multiplicities) is the same for
$$
f(z)-\frac{e^z-1}{z}~~\text{and}~~ \frac1z.
$$
However, the first function has a removable singularity and is thus holomorphic, thus the roots-poles balance is positive, while the second has only one pole, so the balance is negative. By contradiction, the negated claim is wrong.
