For $f \in C^2 : \mathbb{R}^3 \rightarrow \mathbb{R}, s.t. ∆f > 0$ maximum of $f$ on $B(0,r)$ is a strictly increasing function of $r$. For $f \in C^2 : \mathbb{R}^3 \rightarrow \mathbb{R}, s.t.  ∆f > 0$ prove that the maximum of $f$ on $B(0,r)$ is a strictly increasing function of $r$. 
I can take the differtial of $f$ and use the divergence theorem on it, yielding $0 < \int_{B(0,r)}∆f = \int_{\partial B(0,r)}Df \cdot N$, where N is the normal unit vector, or in this case, $\frac{x}{r}$. I feel like this is very close to the answer, because intuitively, it means that $f$ grows on the boundary of the ball, but it doesn't necessarily mean that the maximum of $f$ grows with it. Any tips would be welcome.
 A: Let's take $r_1$ and $r_2$ strictly positive, such as $r_1<r_2$.
 Let's call $B_1=B(0,r_1)$ and $m_1=max_{x\in B_1} f$ exists because $B_1$ is compact. It is the same for $m_2$.
 We know $m_1 \in B_2$ because $r_1<r_2$.
 So $m_1 <= m_2 $ because $m_2$ is the maximum of $B_2$.
 Now we just miss the stricty character.
 Let's suppose it exists $r_1$ and $r_2$ such as $r_1<r_2$ and $m_1=m_2$.
 So all the elements that are in $A=B_2 - (B_1)$  (that we added), are less or equal to $m_1$. 2 cases :
 First we get a local maximum ($m_1$) because we can build a open ball around it where it is the strict maximum. The gradient is equal to zero in $m_1$, and then the Laplace operator is equal to zero (because it is the divergence of the gradient) contradiction with $\Delta f>0$
 Second, every element of $A$ is equal to m1 (because less or equal and not less). We have some areas which are dense where the function is constant, the gradient is equal to zero,and then the Laplace operator, same contradiction.
