Let $u$ be such that $\Delta u =1$ in S=$\{x | |x| <1\}$ with $u=0$ on $\partial S$, then prove that $u(2/3,0) \le 13/18$.
Put $v=u-\frac{|x|^2}{4}$, then $v$ satisfies $\Delta v=0$ in S with $v=-\frac{|x|^2}{4}$ on $\partial S$ i.e. $v= -1/4$ on $\partial S$. Now we know a harmonic function attains its maximum at the boundary, therefore $v \le -1/4 \implies u \le |x|^2/4-1/4 \le 0$ on S, thus the the desired result is always true irrespective of the point. Is it correct?