<s>[INST] Prove that this expression is greater than 1/2 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. 
 [/INST] By the Max Alekseyev's hint we need to prove that $\sum\limits_{cyc}f(a)\geq\frac{3}{4},$ 
where $f(x)=x^2+\frac{2}{\pi^2}\sin^2\pi x,$ $a$, $b$ and $c$ are positives such that $a+b+c=1.$
We have $$f''(x)=4\left(\frac{1}{2}+\cos2\pi x\right),$$
which gives that $f$ is a convex function on $\left[0,\frac{1}{3}\right]$ and on $\left[\frac{2}{3},1\right]$ and $f$ is a concave function on $\left[\frac{1}{3},\frac{2}{3}\right]$.
Now, since it's impossible that $0<a\leq\frac{1}{3}<b\leq\frac{2}{3}<c<1$ and two of variables are placed on $\left(\frac{2}{3},1\right]$,
it's enough to consider two following cases.


*

*$\{a,b\}\subset\left[0,\frac{1}{3}\right]$.


Let $\frac{a+b}{2}=x$.
Thus, by Jensen:
$$\sum_{cyc}f(a)\geq2f\left(\frac{a+b}{2}\right)+f(c)=2f(x)+f(1-2x)=$$
$$=2x^2+\frac{4}{\pi^2}\sin^2\pi x+(1-2x)^2+\frac{2}{\pi^2}\sin^2\pi(1-2x).$$
Id est, it's enough to prove that
$$2x^2+\frac{4}{\pi^2}\sin^2\pi x+(1-2x)^2+\frac{2}{\pi^2}\sin^2\pi(1-2x)\geq\frac{3}{4}$$ or
$$6x^2-4x+\frac{1}{4}+\frac{2}{\pi^2}(2\sin^2\pi x+\sin^22\pi x)\geq0,$$ which is very strong, but true.


*$\{a,b\}\subset\left[\frac{1}{3},\frac{2}{3}\right]$ and $a\geq b$.


Thus, since $\left(a+b-\frac{1}{3},\frac{1}{3}\right)\succ(a,b),$ by Karamata we obtain:
$$\sum_{cyc}f(a)\geq f\left(a+b-\frac{1}{3}\right)+f\left(\frac{1}{3}\right)+f(c)=$$
$$=f\left(\frac{2}{3}-c\right)+f\left(\frac{1}{3}\right)+f(c)=$$
$$=2c^2-\frac{4}{3}c+\frac{5}{9}+\frac{3}{2\pi^2}+\frac{2}{\pi^2}\left(\sin^2\left(\frac{2\pi}{3}-\pi c\right)+\sin^2\pi c\right)\geq\frac{3}{4},$$
where the last inequality is true for all $c\in\left(0,\frac{1}{3}\right).$
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