TITLE: An inequality in Evans' PDE
QUESTION [7 upvotes]: In Section $9.2$ Theorem $5$ of Lawrence Evans' Partial Differential Equations, First Edition the author proves that for a large enough $\lambda$, the equation 
$$\begin{array}-\Delta u+b(\nabla u)+\lambda u=0\ &\mbox{ in } U\\
u=0&\mbox{on }\partial U\end{array}$$
has a solution in $H_0^1(U)$.
On page 507, the author writes 
$$\int_UC(|\nabla u|+1)|u|dx\leq\frac{1}{2}\int_U|\nabla u|^2dx+C\int_U(|u|^2+1)dx\ \mbox{ for }u\in H_0^1(U).$$ 
Here $C$ is the Lipschitz constant for the Lipschitz function $b$.
My problem is that I cannot show this no matter how much I try. How is the gradient term becoming independent of $C$? Could someone please help!

REPLY [6 votes]: Since the OP didn't seem to understand my comment, I'll make it into an answer.  The Peter-Paul inequality (one guy big, the other one small) is the simple arithmetic estimate ($ab\in\mathbb{R}$, $\varepsilon>0$)
$$
ab \le \varepsilon a^2 + \frac1{4\varepsilon}b^2\,.
$$
This simple little guy is all you need to establish the OP's desired estimate, once you accept the fact that the $C$ is different on each side.
In detail, we obtain the estimates
$$
C|u| \le \frac{C^2}2|u|^2 + \frac12\,,
$$
and
$$
C|\nabla u|\,|u| \le \frac12|\nabla u|^2 + \frac{C^2}2|u|^2\,,
$$
which we integrate and add together to conclude
$$
\int_UC(|\nabla u|+1)|u|dx
\leq\frac{1}{2}\int_U|\nabla u|^2dx + C^2\int_U|u|^2dx + \frac12\int_Udx\,,
$$ 
as desired.