Cauchy inequality This is supposed to be an upper bound counterpart for the Cauchy inequality. Let $f$ be entire and  $M(R) = sup_{|z|=R} |f(z)|$ and  $A(R) = sup
_{n≥0} |a_n|R^n$ prove that $2A(2R)$ ≥ $M(R)$
I used Cauchy formula on circle $|z| =2r$ :
                 $M(R) =$ $1/2{\pi}i\int f(z)/(z-z_0) dz $ for some $z_0$ on $|z| = R$ 
Then I tried to compare it with  $2A(2R)$ = $2 sup
_{n≥0} [ 1/2{\pi}i\int f(z) e^{-ni\phi}  d\phi]$
Everything looks quite close, but is not really there. Can someone help me? 

For all $z$, $|z| = R$ implies $|f(z)| \le \sum\limits_{n = 0}^\infty |a_n|R^n \le A(2R)\sum\limits_{n = 0}^\infty \frac{1}{2^n} = 2A(2R)$. Hence $M(R) \le 2A(2R)$.
