Return to Answer

This inequality cannot hold in general. Indeed, \begin{equation*} |\nabla(f^{\frac{p+q-2}{p}})|^p=k^pf^{q-2}|\nabla f|^p, \end{equation*} where \begin{equation*} k:=\frac{p+q-2}p=1+\frac{q-2}p\ge1. \end{equation*} So, at all points where $g>0$, $|\nabla g|>0$, and $|\nabla f|>0$, we can rewrite the inequality in question as \begin{equation*} r-1\le C(s,q)(r^p-k^p) \tag{1}\label{1} \end{equation*} where $r:=a/b$, $a:=f/g$, and $b:=|\nabla f|/|\nabla g|$.
In general, $r$ can take any nonnegative real value.

Letting now $r\to\infty$ in \eqref{1}, we get $C(s,q)>0$. Letting then $r=1$, we get a contradiction: $0\le C(s,q)(1-k)<0$ if $k>1$, that is, if $q>2$.

If, finally, $q=2$, then $k=1$ and the only value of $C(s,q)$ such that \eqref{1} holds for all real $r\ge0$ is $p$ -- so that $C(s,2)$ must depend, not on $s$, but on $p$.

This inequality cannot hold in general. Indeed, \begin{equation*} |\nabla(f^{\frac{p+q-2}{p}})|^p=k^pf^{q-2}|\nabla f|^p, \end{equation*} where \begin{equation*} k:=\frac{p+q-2}p=1+\frac{q-2}p\ge1. \end{equation*} So, at all points where $g>0$, $|\nabla g|>0$, and $|\nabla f|>0$, we can rewrite the inequality in question as \begin{equation*} r-1\le C(s,q)(r^p-k^p) \tag{1}\label{1} \end{equation*} where $r:=a/b$, $a:=f/g$, and $b:=|\nabla f|/|\nabla g|$.
In general, $r$ can take any nonnegative real value.

Letting now $r\to\infty$ in \eqref{1}, we get $C(s,q)>0$. Letting then $r=1$, we get a contradiction: $0\le C(s,q)(1-k)<0$ if $k>1$, that is, if $q>2$.

If, finally, $q=2$, then $k=1$ and the only value of $C(s,q)$ such that \eqref{1} holds for all real $r\ge0$ is $1/p$ -- so that $C(s,2)$ must depend, not on $s$, but on $p$.

This inequality cannot hold in general. Indeed, \begin{equation*} |\nabla(f^{\frac{p+q-2}{p}})|^p=k^pf^{q-2}|\nabla f|^p, \end{equation*} where \begin{equation*} k:=\frac{p+q-2}p=1+\frac{q-2}p\ge1. \end{equation*} So, at all points where $g>0$, $|\nabla g|>0$, and $|\nabla f|>0$, we can rewrite the inequality in question as \begin{equation*} r-1\le C(s,q)(r^p-k^p) \tag{1}\label{1} \end{equation*} where $r:=a/b$, $a:=f/g$, and $b:=|\nabla f|/|\nabla g|$.
In general, $r$ can take any nonnegative real value.

Letting now $r\to\infty$ in \eqref{1}, we get $C(s,q)>0$. Letting then $r=1$, we get a contradiction: $0\le C(s,q)(1-k)<0$ if $k>1$, that is, if $q>2$.

This inequality cannot hold in general. Indeed, \begin{equation*} |\nabla(f^{\frac{p+q-2}{p}})|^p=k^pf^{q-2}|\nabla f|^p, \end{equation*} where \begin{equation*} k:=\frac{p+q-2}p=1+\frac{q-2}p\ge1. \end{equation*} So, at all points where $g>0$, $|\nabla g|>0$, and $|\nabla f|>0$, we can rewrite the inequality in question as \begin{equation*} r-1\le C(s,q)(r^p-k^p) \tag{1}\label{1} \end{equation*} where $r:=a/b$, $a:=f/g$, and $b:=|\nabla f|/|\nabla g|$.
In general, $r$ can take any nonnegative real value.

Letting now $r\to\infty$ in \eqref{1}, we get $C(s,q)>0$. Letting then $r=1$, we get a contradiction: $0\le C(s,q)(1-k)<0$ if $k>1$, that is, if $q>2$.

If, finally, $q=2$, then $k=1$ and the only value of $C(s,q)$ such that \eqref{1} holds for all real $r\ge0$ is $p$ -- so that $C(s,2)$ must depend, not on $s$, but on $p$.

This inequality cannot hold in general. Indeed, at all points where $g>0$, $|\nabla g|>0$, and $|\nabla f|>0$, we can rewrite the inequality in question as \begin{equation} r-1\le C(s,q)(r^p-(p+q-2)) \tag{1}\label{1} \end{equation} where $r:=a/b$, $a:=f/g$, and $b:=|\nabla f|/|\nabla g|$. In general, $r$ can take any nonnegative real value.

Letting now $r\to\infty$ in \eqref{1}, we get $C(s,q)>0$. Letting then $r=1$, we get a contradiction: $0\le C(s,q)(3-(p+q))<0$, since $p,q\ge2$.

This inequality cannot hold in general. Indeed, \begin{equation*} |\nabla(f^{\frac{p+q-2}{p}})|^p=k^pf^{q-2}|\nabla f|^p, \end{equation*} where \begin{equation*} k:=\frac{p+q-2}p=1+\frac{q-2}p\ge1. \end{equation*} So, at all points where $g>0$, $|\nabla g|>0$, and $|\nabla f|>0$, we can rewrite the inequality in question as \begin{equation*} r-1\le C(s,q)(r^p-k^p) \tag{1}\label{1} \end{equation*} where $r:=a/b$, $a:=f/g$, and $b:=|\nabla f|/|\nabla g|$.
In general, $r$ can take any nonnegative real value.

Letting now $r\to\infty$ in \eqref{1}, we get $C(s,q)>0$. Letting then $r=1$, we get a contradiction: $0\le C(s,q)(1-k)<0$ if $k>1$, that is, if $q>2$.