Let $0<x < y < 1$ be given. Prove
$$4x^{2}+4y^{2}-4xy-4y+1 + \frac{4}{\pi^2}\Big[
\sin^{2}(\pi x)+ \sin^{2}(\pi y) + \sin^{2}[\pi(y-x)] \Big] \geq \frac{1}{2}$$


I have been working on this problem for a while now and I have hit a wall. I have plotted this and it seems to be true. I also tried to prove that it is greater than some other fraction such as $\frac{2}{5}$ or something. So far, I have tried doing partial derivative with respect to x but unfortunately, it is very hard to find the roots as it involve equation involving $\sin(2 \pi x)$, $\sin[2 \pi (x-y)]$ and $x$. I think Taylor Series expansion would look very ugly. I would really appreciate it if anyone has any suggestions on what to try or any branch of math I can read on, as I am not sure how to deal with this kind of expression except the elementary tool I have. 



Prove that this expression is greater than 1/2
Let
$$f(x,y):=4x^{2}+4y^{2}-4xy-4y+1 + \frac{4}{\pi^2}\Bigl(
\sin^{2}(\pi x)+ \sin^{2}(\pi y) + \sin^{2}(\pi y-\pi x) \Bigr).$$
I will show that
$$\min\_{0\leq x\leq y\leq 1}f(x,y)=\min\_{0\leq x\leq 1/3}f(x,2x).\tag{$\ast$}$$
This suffices, because the minimum of the one-variable function $f(x,2x)$ is easy to analyze numerically: it occurs around $x\_0\approx 0.146625$, and has value $f(x\_0,2x\_0)\approx 0.502187$.


In order to prove $(\ast)$, observe that $f(x,y)$ is invariant under the following two permutations of the closed triangle $T:=\{(x,y):0\leq x\leq y\leq 1\}$:
$$(x,y)\mapsto (y-x,1-x)\qquad\text{and}\qquad (x,y)\mapsto (x,1-y+x).$$
These two bijections generate a group $G\cong S\_3$ of permutations of $T$. The elements of $G$ are:
\begin{align\*}
&(x,y)\mapsto(x,y),&&(x,y)\mapsto(y-x,1-x),&&(x,y)\mapsto(1-y,1-y+x),\\
&(x,y)\mapsto(x,1-y+x),&&(x,y)\mapsto(1-y,1-x),&&(x,y)\mapsto(y-x,y).
\end{align\*}
The closed triangle $U$ with vertices 
$$A:=(0,0),\qquad B:=(1/2,1/2), \qquad C:=(1/3,2/3)$$ 
is a fundamental domain for the action of $G$ on $T$, hence
$$\min\_T f(x,y)=\min\_U f(x,y).$$
For any interior point $(x,y)\in\mathrm{int\,} U$, we have $0<y<2x<1$ and $0<y<2/3$, therefore
$$\frac{\pi}{4}\cdot\frac{\partial f}{\partial x}=(2\pi x-\pi y)+2\cos(\pi y)\sin(2\pi x-\pi y)>(2\pi x-\pi y)-\sin(2\pi x-\pi y)>0.$$
So there is no local extremum on the interior of $U$, which implies that
$$\min\_T f(x,y)=\min\_{\partial U} f(x,y).$$
It is straightforward to verify that the minimum over the side $AB$ (resp. $BC$) exceeds $0.6$ (resp. $0.57$). Hence the minimum over $\partial U$ equals the minimum over the side $CA$, and $(\ast)$ follows.


