= Solution
Let $M=\|F'\|_\infty$. The sharper <energy estimate> uses the divergence structure of the <viscous scalar conservation law>. Multiply by $u$ and integrate over the line. With $G(s)=\int_0^s rF'(r)\,dr$,
$$
\int u\partial_xF(u)=\int\partial_xG(u)=0,
$$
because the decay makes $G(u)$ tend to zero at both ends. <Integration by parts> in the diffusion term therefore gives
$$
\frac12\frac{d}{dt}\|u(t)\|_2^2+\varepsilon\|u_x(t)\|_2^2=0,
\qquad
\boxed{\|u(t)\|_2^2+2\varepsilon\int_0^t\|u_x(s)\|_2^2\,ds=\|u(0)\|_2^2.}
$$
\b[One may take $C_0=0$, uniformly in $\varepsilon$.] This does not require $F(0)=0$.
If a bound explicitly involving $M$ and $\varepsilon$ is desired, retaining the transport term and using the <Cauchy-Schwarz inequality> and the elementary inequality $ab\leq(a^2+b^2)/2$ gives
$$
M\|u\|_2\|u_x\|_2\leq\frac\varepsilon2\|u_x\|_2^2+\frac{M^2}{2\varepsilon}\|u\|_2^2.
$$
The <Gronwall inequality> then gives the valid but weaker choice $C_0=M^2/\varepsilon$. The exact cancellation explains why its divergence is unnecessary.
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