An inequality in the proof of characterization of the $H^{-1}$ norm in Evans's PDE book

It's an application of Cauchy-Schwarz inequality(ies). Since $u$, $u_{x_i}$ and $g^i$are in $L^2(U)$ we have $\int_ug^0udx\leq \sqrt{\int_U|g^0|^2dx}\sqrt{\int_U|u|^2dx}$ and for all $i\in\{1,\ldots, n\}:\int g^iu_{x_i}\leq \sqrt{\int_U|g^i|^2dx}\sqrt{\int_U|u_{x_i}|^2dx}$, hence \begin{align*} \lVert u\rVert_{H_0^1(U)}^2&=\int_U g^0udx+\sum_{i=0}^ng^iu_{x_i}dx\\ &\leq \sqrt{\int_U|g^0|^2dx}\sqrt{\int_U|u|^2dx}+\sum_{i=1}^n\sqrt{\int_U|g^i|^2dx}\sqrt{\int_U|u_{x_i}|^2dx} \\ &\leq \sqrt{\int_U|g^0|^2dx+\sum_{i=1}^n\int_U |g^i|^2dx}\sqrt{\int_U|u|^2dx+\sum_{i=1}^n\int_U|u_{x_i}|^2dx}, \end{align*} applying this time Cauchy-Schwarz inequality for sums, namely $\displaystyle\sum_{i=0}^n a_ib_i\leq\sqrt{\sum_{i=0}^n|a_i|^2}\sqrt{\sum_{i=0}^n|b_i|^2}$, for $\displaystyle a_0=\sqrt{\int_U|g^0|^2dx}$, $\displaystyle b_0=\sqrt{\int_U|u|^2dx}$, $\displaystyle a_i=\sqrt{\int_U|g^i|^2dx}$ and $\displaystyle b_i=\sqrt{\int_U|u_{x_i}|^2dx}$ for $1\leq i\leq n$. If $u=0$, the inequality we have to show is obvious, and if it's not the case we can divide by $\displaystyle\sqrt{\int_U|u|^2dx+\sum_{i=1}^n\int_U|u_{x_i}|^2dx}$.