I am denoting as $D(z,r)$ the disk in the complex plane, centered at $z$ with radius $r$. Let $h$ be a function harmonic on a neighbourhood of $\overline{D(0,\rho)}$ and for $0\leq r\leq \rho$ define $$M_h(r)\doteqdot \sup_{|z|=r}h(z)$$ The question is to prove that
$$M_h(r)\leq \frac{2r}{\rho+r}M_h(\rho)+\frac{\rho-r}{\rho+r}h(0)\qquad(0\leq r\leq \rho).$$
I could not do much to progress this problem. As I can say, since $h$ is not necessarily positive, tools like Harnack Inequality cannot be used here. So one may only has its hope for using the Poisson-Integral Formula, which states that for $0\leq r\leq \rho$,
$$h(re^{it})=\int_{0}^{2\pi}\frac{\rho^2-r^2}{\rho^2-2\rho r\cos(\theta-t)+r^2}h(\rho e^{i\theta})d\theta\qquad (0\leq t\leq 2\pi).$$
Even so, I cannot see what estimations I may use so that even the first term in the right hand side in the above inequality can be obtained. In particular, I cannot see how such denominator appears.
I am thanking beforehand anyone who wishes to help me figure out this exercise.
Best regards.
I wish to personally thank user10354138 for claryfying this question, so that we have this answer.
Since $h$ is harmonic on a neighbourhood of $\overline{D(0,\rho)}$, we have that $$M_\rho (h)=M<+\infty.$$ In particular, the function $$M-h\text{ is harmonic in a neighbourhood of }\overline{D(0,\rho)}\,.$$ It is also positive on the disk $D(0,\rho)$. Hence, it justfies the use of Harnack's Inequality and so $$(M_\rho-h)(re^{it})\geq \frac{\rho-r}{\rho+r}(M_\rho-h)(0)\qquad(0\leq r<\rho)\,.$$ After some computations, we conclude that $$h(re^{it})\leq \frac{2r}{\rho+r}M_\rho+\frac{\rho-r}{\rho+r}h(0)$$ Taking now the supremum over all $t\in[0,2\pi]$, we reach to a conclusion. That is, $$M_r(h)\leq \frac{2r}{\rho+r}M_\rho+\frac{\rho-r}{\rho+r}h(0)\,.$$ The case $r=\rho$ is trivialy justified.