Minimal surfaces maximum principle This is homework so no answers please.
The problem is
The domain is unit disk in $\mathbb{R}^{2}$
Suppose u,w satisfy the minimal-surface equation $div(\frac{\nabla u}{\sqrt{1+|\nabla u|^2}})=0$, show that their difference satisfies an elliptic equation; by satisfies I mean solution, subsolution or supersolution ie. $L(u-w)=0, \leq 0$ or $\geq 0$
Attempts(You can ignore)

1)In dimension 2 we have $div(\frac{\nabla u}{\sqrt{1+|\nabla u|^2}})=0\Rightarrow (1+u^{2}_{x})u_{yy}-2u_{x}u_{y}u_{xy}+(1+u^{2}_{y})u_{xx}=0$. 
So after calculutations... I am done, if I could show $B:=(\frac{1}{\sqrt{1+u_{x}^{2}+u_{y}^{2}}}-\frac{1}{\sqrt{1+w_{x}^{2}+w_{y}^{2}}})(u_{y}+w_{x})$ satisfies any of the following:
$$div(B)=0, \leq 0~ or~ \geq 0$$
this is a strong claim so I doubt it will work.
2)Let $G(v):=\frac{  v}{(1+| v|^{2}){\frac{1}{2}}}$, then by FTC
$G(\bigtriangledown v)-G(\bigtriangledown w)=\int_{0}^{1}\frac{d}{dt}G(\bigtriangledown v +t (\bigtriangledown w -\bigtriangledown v))dt=$
by chain rule and since $(\bigtriangledown w -\bigtriangledown v)$ independent of t
$=[\int_{0}^{1}dG(\bigtriangledown v +t (\bigtriangledown w -\bigtriangledown v))dt] (\bigtriangledown w -\bigtriangledown v)=:(a^{ij})(\bigtriangledown w -\bigtriangledown v).$
But then I don't know how to show $[\int_{0}^{1}dG(\bigtriangledown v +t (\bigtriangledown w -\bigtriangledown v))dt] $ is a positive definite matrix

Any suggestions?
Thank you
Update
I showed that $div(\frac{\nabla u}{\sqrt{1+|\nabla u|^2}})=0$ is an Elliptic PDE, does that tell me anything about u-w?
 A: For v,w having minimal graphs, the goal is to show $v-w$ satisfies an elliptic equation in order to apply the Strong Maximum principle. Our minimal surface equation is $\sum_{j=1}^{2}D_{i}(\frac{  D_{i}v}{(1+| D v|^{2}){\frac{1}{2}}}))=:\sum_{j=1}^{2}D_{i}(  G_{i}(D v) )$, where $ G_{i}(v):=\frac{  v_{i}}{(1+| v|^{2}){\frac{1}{2}}}$. 
Therefore, we can write $0=\sum_{i=1}^{2}D_{i}(  G_{i}(D v)-G_{i}(D w) )$. The goal is to show that $G_{i}(D v)-G_{i}(D w) $ can be writen in the form $\sum_{j=1}^{2}a^{ij}(Du,Dw)D_{i}(u-w)$ where $a^{ij}$ is an elliptic coefficient and so by linearity of derivative we get the elliptic pde in divergence form
$$0=\sum_{i=1}^{2}D_{i}(  G_{i}(D v)-G_{i}(D w) )=\sum_{i=1}^{2}D_{i}(\sum_{j=1}^{2}a^{ij}(Du,Dw)D_{i}(u-w))=$$
$$=\sum_{i=1}^{2}\sum_{j=1}^{2}D_{i}(a^{ij}(Du,Dw)D_{i}(u-w)).$$
Then by the Strong maximum principle $u-w\leq min_{D}(u-w)=min_{\partial D}(u-w)\geq 0$.\
so we start with FTC for the $G_{i}(D v)-G_{i}(D w)$
$$G_{i}(D v)-G_{i}(D w)=\int_{0}^{1}\frac{d}{dt}G_{i}(D v +t (D w -D v))dt=$$
by chain rule we get the factor $(D w -D v)$
$$=[\int_{0}^{1}dG_{i}(D v +t (D w -D v))dt]\cdot (D w -D v),$$
in order to compute the Jacobian f $G_{i}$ we write $dG_{i}(x)=(\frac{\partial}{\partial x_{1}}G_{i}(x),\frac{\partial}{\partial x_{2}}G_{i}(x))$ where 
$x:=D v +t (D w -D v))$; in other words, for fixed t we treat $D v +t (D w -D v))$ as a variable in $\mathbb{R}^{2}$. Thus, the dot product of the Jacobian and $(D w -D v)$ equals
$$=\sum_{j=1}^{2} a_{ij}(Dv,Dw)(D_{j}v-D_{j}w),$$
where $a_{ij}(Dv,Dw):=\int_{0}^{1}\frac{\partial }{\partial x_{j} }G_{i}(x)dt$ and x as above. So now it remains to show that $A:=(a_{ij}(Dv,Dw))$ is a positive definite matrix. \
We have  $G_{i}(x)=\frac{  x_{i}}{(1+| x|^{2})^{\frac{1}{2}}}=\frac{ \partial}{\partial  x_{i}} (1+| x|^{2})^{\frac{1}{2}}$ and so $a_{ij}(Dv,Dw):=\int_{0}^{1} \frac{\partial^{2} }{\partial x_{j}\partial x_{i} }(1+| x|^{2})^{\frac{1}{2}}dt$ i.e. the matrix A is
$$A=\bigl(\begin{smallmatrix}
\int_{0}^{1} \frac{\partial^{2} }{\partial x_{1}^{2} }(1+| x|^{2})^{\frac{1}{2}}dt &\int_{0}^{1} \frac{\partial^{2} }{\partial x_{1}\partial x_{2} }(1+| x|^{2})^{\frac{1}{2}}dt \\ 
\int_{0}^{1} \frac{\partial^{2} }{\partial x_{1}\partial x_{2} }(1+| x|^{2})^{\frac{1}{2}}dt & \int_{0}^{1} \frac{\partial^{2} }{\partial x_{2}^{2} }(1+| x|^{2})^{\frac{1}{2}}dt
\end{smallmatrix}\bigr)=\int_{0}^{1}\bigl(\begin{smallmatrix}
 \frac{\partial^{2} }{\partial x_{1}^{2} }(1+| x|^{2})^{\frac{1}{2}} & \frac{\partial^{2} }{\partial x_{1}\partial x_{2} }(1+| x|^{2})^{\frac{1}{2}} \\ 
 \frac{\partial^{2} }{\partial x_{1}\partial x_{2} }(1+| x|^{2})^{\frac{1}{2}} &  \frac{\partial^{2} }{\partial x_{2}^{2} }(1+| x|^{2})^{\frac{1}{2}}
\end{smallmatrix}\bigr)dt.$$
So it suffices to show that  $[\int_{0}^{1}\frac{\partial }{\partial x_{j} }B_{t}dt]$ is positive definite if $B_{t}$ is for all t. By positive definiteness $\left \langle \xi, B_{t} \right \rangle=\sum_{ij}(B_{t})_{ij}x_{i}x_{j}>0$. Integrating a positive quantity is positive i.e. $\int_{0}^{1}\left \langle \xi, B_{t} \right \rangle dt>0$. Then by linearity of integral  $\left \langle \xi, \int_{0}^{1}B_{t}dt \right \rangle=\int_{0}^{1}\left \langle \xi, B_{t} \right \rangle dt>0$.
