Prove that this expression is greater than 1/2

Question

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.

Answer 1

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.

Answer 2

Let $F(x,y)$ denote the left-hand side of your inequality. It is easy to see that $|\nabla F(x,y)|\sqrt2/n<0.002$ if $0<x<y<1$, where $n:=6600$. A direct calculation shows that $F(i/n,j/n)>0.5021\dots$ for all integers $i,j$ such that $0\le i\le j\le n$. It follows that $F(x,y)>0.502-0.002=1/2$ if $0<x<y<1$, as desired.

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Details of the calculations can be seen in the image of a Mathematica notebook below. Note that Mathematica can compute any expression in elementary functions with any degree of accuracy.

One can see that the calculation of the minimum of $F(i/n,j/n)$ over all integers $i,j$ such that $0\le i\le j\le n=6600$ took about 422 sec (on 12 cores working in parallel). Without parallelization, the execution time was about 2290 sec.

It is also seen that the infimum of $F(x,y)$ over all $x,y$ such that $0<x<y<1$ is $<0.5022$, which is pretty close to $1/2$.

Answer 3

$\newcommand{\R}{\mathbb{R}}$ An advantage of my previous answer was that, while the computer calculations were pretty heavy there, the logic was extremely simple; virtually no thinking or ingenuity was needed.

On the other hand, one can use a bit of thinking in order to greatly reduce the amount of calculations. More specifically, one can use second-order partial derivatives instead of the first-order ones, based on the following simple

Lemma 1 Suppose that a function $F\colon[x_1,x_2]\times[y_1,y_2]\to\R$ is such that for some real $c$ and $M$ we have $F(x_i,y_j)\ge c$ for $i,j$ in $\{1,2\}$, and the second-order partial derivatives $F''_{xx}$ and $F''_{yy}$ are bounded from above by $M$. Suppose also that $x_2-x_1=y_2-y_1=h>0$. Then $$ F\ge c-Mh^2/4 $$ (on $[x_1,x_2]\times[y_1,y_2]$).

It is easy to see that for $F(x,y)$ denoting the left-hand side of your inequality we have $F''_{xx}\le M$ and $F''_{yy}\le M$, where $M:=32$. Hence, for $n=64$ and $h:=1/n$ we have $Mh^2/4<0.002$. One the other hand, a direct calculation shows that $F(i/n,j/n)>c:=0.5023\dots$ for all integers $i,j$ such that $0\le i\le j\le n$. (This calculation now takes about 0.1 sec -- which may be compared with 2290 sec for the previous calculation, based on first-order partial derivatives, with $n=6600$.)

Thus, by Lemma 1, $F(x,y)>c-Mh^2/4=0.5023-0.002>1/2$ if $0<x<y<1$, as desired.

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It remains to prove Lemma 1. Let us first establish its one-dimensional analogue:

Lemma 2 Suppose that a function $f\colon[x_1,x_2]\to\R$ is such that for some real $c$ and $M$ we have $f(x_i)\ge c$ for $i\in\{1,2\}$, and $f''\le M$. Suppose also that $x_2-x_1=h>0$. Then $$ f\ge c-Mh^2/8 $$ (on $[x_1,x_2]$).

Proof of Lemma 2 Let $g(x):=f(x)+M(x-x_1)(x_2-x)/2$. Then $g(x_i)=f(x_i)\ge c$ for $i\in\{1,2\}$ and $g''=f''-M\le0$, so that $g$ is concave and hence $g\ge c$ (on $[x_1,x_2]$). Thus, $f(x)=g(x)-M(x-x_1)(x_2-x)/2\ge c-Mh^2/8$, as claimed. $\Box$

Now we are ready for

Proof of Lemma 1 Take any $(x_*,y_*)\in[x_1,x_2]\times[y_1,y_2]$. For each $j\in\{1,2\}$, applying Lemma 1 to the function $x\mapsto f_j(x):=F(x,y_j)$ in place of $f$, we get $F(x_*,y_j)\ge c_1:=c-Mh^2/8$. Applying now Lemma 1 to the function $y\mapsto g(y):=F(x_*,y)$ in place of $f$, we get $F(x_*,y_*)\ge c_1-Mh^2/8=c-Mh^2/4$, as claimed. $\Box$

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Remark In the answer by GH from MO, the minimization of the function $F$ of two arguments was reduced to the minimization of a function of one argument. Using that reduction together with Lemma 2 above, it should be possible to further reduce the execution time from 0.1 sec to something like 0.1/20=0.05 sec.

Answer 4

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.

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

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.

2. $\{a,b\}\subset\left[\frac{1}{3},\frac{2}{3}\right]$ 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 $c\in\left(0,\frac{1}{3}\right).$

Answer 5

Remarks: In @GH from MO's very nice answer, we need to prove that, for all $x \in [0, 1/3]$, $$f(x, 2x) = 12x^2 - 8x + 1 + \frac{4\left(2\sin^2(\pi x) + \sin^2(2\pi x)\right)}{\pi^2}\ge \frac12 . \tag{1}$$ Here is a proof.

The desired inequality (1) is written as $$12x^2 - 8x + \frac12 + \frac{9}{\pi^2} \ge \frac{\left(3 - 4\sin^2(\pi x)\right)^2}{\pi^2}. \tag{2}$$

Fact 1: It holds that, for all $x\in [0, 1/3]$, $$3 - 4\sin^2(\pi x) \le 3 - 4\sin^2(\pi/7) - 4\pi\sin(2\pi/7)\cdot (x - 1/7).$$ (Proof: Let $f(x) = \mathrm{RHS} - \mathrm{LHS}$. We have $f(1/7) = 0$. We have $f'(x) = 4\pi (\sin (2\pi x) - \sin (2\pi/7))$. Thus, $f'(x) < 0$ on $(0, 1/7)$, and $f'(x) > 0$ on $(1/7, 1/3]$. Thus, we have $f(x)\ge 0$ on $[0, 1/3]$.)

By Fact 1 and (2), it suffices to prove that $$12x^2 - 8x + \frac12 + \frac{9}{\pi^2} \ge \frac{\left(3 - 4\sin^2(\pi/7) - 4\pi\sin(2\pi/7)\cdot (x - 1/7)\right)^2}{\pi^2}$$ which is true (a quadratic inequality).

We are done.