I've founding some difficulties in proving the following statement:
Let $u\in C^2(\mathbb{R}^2)$ subharmonic (i.e. $-\Delta u\leq 0$). For $r\geq 0$ set: $$ M(r)=\max_{x\in\partial B_r(0)}u(x) $$
Prove that for any $0<r_1<r<r_2$: $$ M(r)\leq\dfrac{\log r_2-\log r}{\log r_2-\log r_1}M(r_1)+\dfrac{\log r-\log r_1}{\log r_2-\log r_1}M(r_2) $$
Proof: Let $r=|x|$ and fix $0<r_1<r<r_2$, the function: $$ g(r)=\dfrac{\log r_2-\log r}{\log r_2-\log r_1}M(r_1)+\dfrac{\log r-\log r_1}{\log r_2-\log r_1}M(r_2) $$ is harmonic on the annulus $0<r_1<r<r_2$ because we can rewrite $g(r)=\alpha\log r$ and $\log|x|$ is the fundamental solution to the Laplace equation on $\mathbb{R}^2$. Now the function $u(x)$ is dominated on the boundaries of the annulus $0\leq r_1\leq r\leq r_2$ by $g$, indeed for $x\in\partial B_{r_1}(0)$ we have: $$ u(x)\leq g(r_1)=M(r_1)=\max_{\partial B_{r_1}(0)}u(x) $$ and also for $x\in\partial B_{r_2}(0)$ we have: $$ u(x)\leq g(r_2)=M(r_2)=\max_{\partial B_{r_2}(0)}u(x) $$ By the maximum principle applied to the subharmonic function $u(x)-g(|x|)$ we obtain that: $$ u(x)\leq g(r)=\dfrac{\log r_2-\log r}{\log r_2-\log r_1}M(r_1)+\dfrac{\log r-\log r_1}{\log r_2-\log r_1}M(r_2) \ \ \ 0< r_1< r=|x|< r_2 $$ this implies that we can take the maximum on the LHS for $0<r_1<r<r_2$ and we get: $$ \max_{\partial B_r(0)}u(x)=M(r)\leq\dfrac{\log r_2-\log r}{\log r_2-\log r_1}M(r_1)+\dfrac{\log r-\log r_1}{\log r_2-\log r_1}M(r_2) $$
Is it fine?