Show that for $f$ analytic in $B(0,2)$, $\max_{|z|=1}|\frac{1}{z}-f(z)|\ge 1$? Let $f:B(0,2)\to \Bbb C$ be an analytic function. Show that $$\max_{|z|=1}\left|\frac{1}{z}-f(z)\right|\ge 1.$$
I tried to write $f(z)$ as power series since it is analytic, it doesn't seem work. I also tried to use maximum modulus principle, it also did not work.
Could anyone kindly help? I am really struggling with this problem. Thank you!

Hint: Try integrating $1/z-f(z)$ around the unit circle.
