Let $f$ holomorphic function on$\{z|1\leq |z|\leq 2\}$ If $\operatorname{Re} f>0$ on $\{|z|=2\}$, there exist $z$ s.t., $|z|=1$ and $\operatorname{Re} f\geq 0.$
My idea: if $\operatorname{Re}f<0$ on $\{|z|=1\}$, there exist closed curve $\gamma$ in $\{z|1\leq |z|\leq 2\}$ s.t., $\operatorname{Re}f=0$ on $\gamma$. But I can't make contradiction.