[@DiB:dpe], $$\begin{gathered}
    \label{eq:monot_i}
    \big(
    {\lvert{D}_{y}{u}(x)\rvert}^{p-2}
    {D}_{y}{u}(x)
    -
    {\lvert{D}_{y}{v}(x)\rvert}^{p-2}
    {D}_{y}{v}(x)
    \big)
    {D}_{y}({u}(x)-{v}(x))
    \,
    \mathcal{A}
    \\
    \ge
    \gamma_{0}(p)
    {\lvert{D}_{y}({u}(x)-{v}(x))\rvert}^{p}
    \,
    \mathcal{A}
    \,,
  \end{gathered}$$ where we define $$\mathcal{A}
    =
    \frac{
      {D}_{y}{{\ifthenelse{\equal{*}{*}}{({u}(x)-{v}(x))_{+}}{\left({u}(x)-{v}(x)\right)_{+}}}}^{q}
    }{
      {D}_{y}({u}(x)-{v}(x))
    }
    \ge 0
    \,.$$

On the other hand, we write the right hand side of with $h=0$ as $$\label{eq:monot_ii}
    {\lvert{D}_{y} ({u}(x)-{v}(x))\rvert}^{p}
    \,
    \mathcal{B}
    \,,
    \qquad
    \mathcal{B}
    :=
    {\left|
      \frac{
        {D}_{y}{{\ifthenelse{\equal{*}{*}}{({u}(x)-{v}(x))_{+}}{\left({u}(x)-{v}(x)\right)_{+}}}}^{\frac{q-1+p}{p}}
      }{
        {D}_{y}({u}(x)-{v}(x))
      }
    \right|}^{p}
    \,.$$ Therefore we have only to prove that $\mathcal{A}\ge \gamma_{0}\mathcal{B}$. Clearly in doing so we may assume without loss of generality that $${u}(y)
    -
    {v}(y)
    >
    {u}(x)
    -
    {v}(x)
    \,.$$ Hence it is left to prove that $$\begin{gathered}
    \label{eq:monot_iii}
    \big[
    {u}(y)-{v}(y)
    -
    ({u}(x)-{v}(x))
    \big]^{p-1}
    \big[
    {{\ifthenelse{\equal{*}{*}}{({u}(y)-{v}(y))_{+}}{\left({u}(y)-{v}(y)\right)_{+}}}}^{q}
    -
    {{\ifthenelse{\equal{*}{*}}{({u}(x)-{v}(x))_{+}}{\left({u}(x)-{v}(x)\right)_{+}}}}^{q}
    \big]
    \\
    \ge
    \gamma_{0}
    \Big[
    {{\ifthenelse{\equal{*}{*}}{({u}(y)-{v}(y))_{+}}{\left({u}(y)-{v}(y)\right)_{+}}}}^{\frac{q-1+p}{p}}
    -
    {{\ifthenelse{\equal{*}{*}}{({u}(x)-{v}(x))_{+}}{\left({u}(x)-{v}(x)\right)_{+}}}}^{\frac{q-1+p}{p}}
    \Big]^{p}
    \,.
  \end{gathered}$$ Denote $$a
    =
    {u}(y)
    -
    {v}(y)
    \,,
    \qquad
    b
    =
    {u}(x)
    -
    {v}(x)
    \,.$$ If $b\le 0$, is obviously satisfied with $\gamma_{0}=1$. If $b>0$, by Hölder’s inequality we have $$\begin{gathered}
    \label{eq:monot_iv}
    \big[
    a^{\frac{q-1+p}{p}}
    -
    b^{\frac{q-1+p}{p}}
    \big]^{p}
    =
    \Big[
    \frac{q-1+p}{p}
    \int_{b}^{a}
    s^{\frac{q-1}{p}}
    {\,\textup{\textmd{d}}}s
    \Big]^{p}
    \\
    \le
    \gamma(q,p)
    \Big[
    \int_{b}^{a}
    s^{q-1}
    {\,\textup{\textmd{d}}}s
    \Big]
    \Big[
    \int_{b}^{a}
    {\,\textup{\textmd{d}}}s
    \Big]^{p-1}
    \le
    \gamma(q,p)
    (a^{q}-b^{q})
    (a-b)^{p-1}
    \,,
  \end{gathered}$$ proving and concluding the proof.

In the following all radii of balls in $G$ will be 