Let $\Omega \subset \mathbb{R}^n$ be an bounded domain with smooth boundary $\partial \Omega$ and consider the functional $I:H_0^{1}(\Omega) \to \mathbb{R}$ defined by
$$I(u)=\frac{1}{2}\lVert u \rVert_{H_0^{1}(\Omega)}^2+\int_{\Omega} u \cos(u)dx-\int_{\Omega} \sin(u)dx, $$
there exist some $u_0$ such that $I(u_0)<0$?
Context of the problem: I was able to prove the existence of weak solution of the partial differential equation
$$-\Delta u+u=u\sin(u)\,\,\,\text{in}\,\,\, \Omega\\ u=0\,\,\,\text{in}\,\,\, \partial \Omega $$ by finding $v$ such that $$I(v)=\inf_{u \in H_0^1{(\Omega})} I(u) $$
I want to show that the weak solution founded is not trivial, to do this, it is enough to find some point in which $I$ is negative.
No, such a function $u_0$ does not exist. Indeed, we have $$ I(u) \ge \int_{\Omega} \frac{|u(x)|^2}{2} + u\cos(u(x)) - \sin(u(x)) dx, $$ so it is enough to show that $$ f(x) := \frac{x^2}2 + x\cos(x) - \sin(x) \ge 0 $$ for all $x\in\Bbb R$ in order to prove that $I(u) \ge 0$ for all $u \in H^1_0(\Omega)$.
Since $f'(x) = x - x\sin(x)$, the critical points of $f$ are exactly at $x = 0$ and $x = \frac{(4n+1)\pi}2$ for all $n\in \Bbb Z$. You can then show that the global minimum of $f$ is at $x=0$ where $f(0) = 0$ since $f$ is increasing on $[0,\infty)$ and decreasing on $(\infty,0]$. This shows that $f \ge 0$ on $\Bbb R$ and hence $I \ge 0$ on $H^1_0(\Omega)$.