The Timoshenko beam theory yields the following variational problem. For $t\in(0,1), f\in L_2$ given, we search for a $(w,\beta)\in V\subset[H^1(0,1)]^2$ such that it is a solution of
\begin{align*} A((w,\beta),(v,\delta)) := \int\limits_{0}^1 \beta'\delta'dx+ \frac{1}{t^2}\int\limits_0^1(w'-\beta)(v'-\delta)dx = \int\limits_0^1 f v dx \ \forall(v,\delta)\in V \end{align*}
I want to prove that the bilinear form $A((w,\beta),(v,\delta))$ is non-negative.
Does anyone know how to do this? I think it might be possible by using Young's inequality.
As said in my comment, $A$ is non-negative because $$A((w,\beta),(w,\beta))=\|\beta'\|_{L^2}^2 +\frac{1}{t^2}\|w'-\beta\|^2_{L^2}\geq 0,\quad\forall \ (w,\beta)\in V.$$
I suspect that you want to prove that $A$ is $V$-continuous. If so, you should use the Hölder's inequality:
Since $$A((w,\beta),(v,\delta)) = \int\limits_{0}^1 \beta'\delta'\;dx+ \frac{1}{t^2}\int\limits_0^1(w'v'-w'\delta-\beta v'+\beta\delta)\;dx,$$ we have \begin{align*} |A((w,\beta),(v,\delta))| &\leq \int\limits_{0}^1 |\beta'||\delta'|\;dx+ \frac{1}{t^2}\int\limits_0^1|w'||v'|+|w'||\delta|+|\beta||v'|+|\beta||\delta|\;dx \end{align*} and thus, by Hölder's inequality, \begin{align*} |A((w,\beta),(v,\delta))| &\leq \|\beta'\|_{L^2}\|\delta'\|_{L^2}+\frac{1}{t^2}\left(\|w'\|_{L^2}\|v'\|_{L^2}+\|w'\|_{L^2}\|\delta\|_{L^2}+\|\beta\|_{L^2}\|v'\|_{L^2}+|\beta\|_{L^2}\|\delta\|_{L^2}\right)\\ &\leq \frac{1}{t^2}\left(\|\beta'\|_{L^2}\|\delta'\|_{L^2}+\|w'\|_{L^2}\|v'\|_{L^2}+\|w'\|_{L^2}\|\delta\|_{L^2}+\|\beta\|_{L^2}\|v'\|_{L^2}+|\beta\|_{L^2}\|\delta\|_{L^2}\right). \end{align*} Note that $\|\beta'\|^2_{L^2}\leq\|\beta\|_{H^1}^2\leq \|(w,\beta)\|_V^2$ and thus $\|\beta'\|_{L^2}\leq \|(w,\beta)\|_V$. A similar estimate is valid for the other terms. Therefore, \begin{align*} |A((w,\beta),(v,\delta))| &\leq \frac{5}{t^2}\|(w,\beta)\|_V\|(v,\delta)\|_V. \end{align*}