Set $$\phi (x) = u(x)+iv(x)$$, $$x=(x_1,\ldots,x_N)$$, a $$T$$-periodic function in $$H^1_\text{loc}(\mathbb{R}^N)$$, that is $$\phi (x) = \phi (x_1 + T ,\ldots, X_N + T)$$ for all $$x$$ and where $$u = \operatorname{Re} \phi$$ and $$v = \operatorname{Im} \phi$$. I wonder if $$\int_{[0,T]^N}\frac{u_{x_1}^2}{2} + \frac{v_{x_1}}{\sqrt{2}}(u-1)^2\ dx > 0$$ for functions satisfying $$\int_{[0,T]^N} v_{x_1}(u-1)\ dx = \varepsilon > 0$$ for some small and fixed $$\varepsilon > 0$$. Any idea or help to show this would be appreciate! I am stucked. Thank you in advance.

EDIT: maybe proving $$\int_{[0,T]^N}\frac{|\nabla u|^2}{2} + \frac{v_{x_1}}{\sqrt{2}}(u-1)^2\ dx > 0$$ under the same constraint is easier?

Take $$N=1$$, $$T=1$$, and let $$u=1/2$$. Let $$v(x) =1/2 - x$$ for $$x\in \mathbb{Z}+[h,1-h]$$ for some small positive number $$h$$. By taking $$h$$ small enough, you can obviously make sure
$$\int_{[0,1]} v'(u-1) = \int_h^{1-h} (-1)(1/2-1) = \dfrac{1}{2}(1-2h)$$
$$\int_{[0,1]} v'(u-1)^2 = \int_h^{1-h}(-1)(1/2-1)^2 = -\dfrac{1}{4}(1-2h)$$
is then negative (and equal to $$\sqrt{2}$$ times your first integral).
You definitely seem to need additional hypotheses on $$u$$ and $$v$$ to have chance of anything like what you propose being true (perhaps you missed some?)