= Solution
For $u,w\in X$, the forcing terms cancel, giving
$$
\begin{aligned}
\langle A(u)-A(w),u-w\rangle
={}&\int_\Omega\bigl(|\nabla u|^2\nabla u-|\nabla w|^2\nabla w\bigr)\cdot(\nabla u-\nabla w)\,dx\\
&+\int_\Omega(u^7-w^7)(u-w)\,dx.
\end{aligned}
$$
The scalar <function> $s\mapsto s^7$ is strictly increasing on all of $\mathbb R$, so the second <integrand> is nonnegative. For the first one, use the <quartic gradient monotonicity inequality>. Put $d=\xi-\eta$ and $m=(\xi+\eta)/2$ for real <vectors> $\xi,\eta$. Expanding gives
$$
\begin{aligned}
(|\xi|^2\xi-|\eta|^2\eta)\cdot(\xi-\eta)
&=|m|^2|d|^2+2(m\cdot d)^2+\frac14|d|^4\\
&\geq\frac14|\xi-\eta|^4.
\end{aligned}
$$
Consequently
$$
\boxed{\langle A(u)-A(w),u-w\rangle
\geq\frac14\|\nabla(u-w)\|_4^4
=\frac14\|u-w\|_X^4>0\quad(u\ne w).}
$$
The last strict inequality uses the <Poincaré inequality>: if the <weak gradient> of $u-w\in W_0^{1,4}(\Omega)$ vanishes, then $u=w$. This proves that $A$ is a <strictly monotone operator>.
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