Prove $\int_0^1 ((g'(x))^2-1)^2dx \geq 1$ for smooth $g$ with $g(0)=g(1)=0$ This came up in an optimization problem.
How do you prove that $\int_0^1 ((g'(x))^2-1)^2dx \geq 1$ for any $g$ which is twice continuously differentiable on $[0,1]$ and such that $g(0)=g(1)=0$?

The functional $\int_0^1 ((g'(x))^2-1)^2dx  $ can be made arbitrarily close to $0$ subject to the stated conditions $g(0)=g(1)=0$ and $g\in C^2$. 
Begin with the triangle $x\mapsto \frac12- |x-1/2|$; formally, the functional is zero for this function but of course this is not a differentiable function. Redefine it in $\epsilon$-neighborhood of $1/2$, replacing the vertex of triangle with a smooth cap. This can be done so that $|g'|$ does not exceed $1$. Hence, 
$$\int_0^1 ((g'(x))^2-1)^2dx  \le \int_{1/2-\epsilon}^{1/2+\epsilon} 1\, dx  =2\epsilon$$ 
