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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.

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  • $\begingroup$ Can you show me some hint? @MarkSapir $\endgroup$ – Seaweed Jul 19 at 3:37
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    $\begingroup$ The inequality can be rewritten as $$f(x) + f(y-x) + f(1-y)\geq \frac{3}{2},$$ where $f(x) := 2x^2 + \frac{4}{\pi^2}\sin(\pi x)^2$. $\endgroup$ – Max Alekseyev Jul 19 at 10:52
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    $\begingroup$ Please don't close this problem as it has nice features (cf. my response and comment below). $\endgroup$ – GH from MO Jul 19 at 15:48
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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.

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  • $\begingroup$ Can I ask how you computed the fundamental domain for the action of G on T? This is probably an uninteresting question and I really want to understand it but I only have undergrad intro to Abstract Algebra. $\endgroup$ – Seaweed Jul 19 at 13:51
  • $\begingroup$ Does this have something to do with Moduli space? $\endgroup$ – Seaweed Jul 19 at 14:11
  • $\begingroup$ @SearidangPa: I discovered the group of functional equations $G$ experimentally. The elements of $G$ are affine linear transformations fixing $T$, hence they fix the baricenter of $T$. Then I calculated the orbit of a vertex of $T$, and observed that it consists of the vertices and midpoints of $T$. From this I figured that the orbit of $U$ consists of the six closed triangles into which the medians cut $T$. This proves that $U$ is a fundamental domain. Let me add that I don't see a connection to moduli space, but perhaps the source of the problem would better explain $G$ and $U$. $\endgroup$ – GH from MO Jul 19 at 15:44
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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.


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 $1/2$.

enter image description here

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    $\begingroup$ A direct calculation shows that F(i/n,j/n)>0.5021… for all integers i,j such that 0≤i≤j≤n. You are amazingly fast at arithmetic! It would take me a few centuries to verify that (the fastest route would be to learn mining, metallurgy, electric engineering, etc., then to build a computer from scratch, so that I would know what it is doing, and then run a program on it). In other words, I will never know in my life if this computation is actually correct though I do not have any grounds to refute it either :-) $\endgroup$ – fedja Jul 24 at 0:33
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    $\begingroup$ @fedja : Very interesting! So, you would not accept any aid of any computer or any calculator, unless you yourself built it? As for me, I'd much rather rely on Mathematica, tested an innumerable number of times by an innumerable number of people, than either on my calculations by hand or on the calculations of any machine I might ever build. :-) Am I wrong in this? $\endgroup$ – Iosif Pinelis Jul 24 at 4:36
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    $\begingroup$ Mathematica is known to make errors too :-). My point of view is simply that one cannot present or accept something as a "proof" unless he understands every little detail on it and the words "direct computation" always mean "a trivial exercise for the reader in elementary algebra". Of course, I'll accept any aid from any source when building the proof, but I'm the only one who is responsible for the final product and have no right to shift this responsibility anywhere else, so I'd better rely on nobody and nothing there. Isn't that how the mathematical writing works? ;-) $\endgroup$ – fedja Jul 24 at 13:07
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    $\begingroup$ Mathematica is known to make errors too :-). I think the key word here is "too". I think of a brain as a machine, too, and on my brain I would rely much less than on Mathematica, at least as far as routine calculations are concerned. Also, my other answer here gives one a chance to check everything by hand (and possibly even more of such a chance by engaging higher-order derivatives), if such a desire is there. :-) $\endgroup$ – Iosif Pinelis Jul 24 at 18:34
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    $\begingroup$ In fact, why should I accept as "a proof" my own computation by a pencil on a sheet of paper, more than the computation of a computer, just because I don't know how the computer works? I don't know how my brain works either, after all. $\endgroup$ – Pietro Majer Jul 28 at 9:10
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$\newcommand{\R}{\mathbb{R}}$ 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.


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. $\Box$


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.

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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.

  1. $\{a,b\}\subset\left[\frac{1}{3},\frac{2}{3}\right]$ 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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  • $\begingroup$ I suspect that the whole difficulty of the problem lies in the part covered by the sentence "which is very strong but true". Indeed, it is a nasty inequality to prove by hand and currently I do not have any decent solution myself. Still, if we figure it out, we have some chance to handle the general version as well... $\endgroup$ – fedja Jul 28 at 22:14
  • $\begingroup$ @fedja I also have no a solution of this inequality by hand. $\endgroup$ – Michael Rozenberg Jul 29 at 3:43
  • $\begingroup$ Note that $6x^2-4x+\frac{1}{4}+\frac{2}{\pi^2}(2\sin^2\pi x+\sin^22\pi x) = \frac{1}{2} [f(x, 2x) - \frac{1}{2}]$. GH from MO's solution does not focus on how to prove that $f(x, 2x) \ge \frac{1}{2}$ or $6x^2-4x+\frac{1}{4}+\frac{2}{\pi^2}(2\sin^2\pi x+\sin^22\pi x)\ge 0$. It seems not easy to come up with a nice solution. $\endgroup$ – River Li Jul 29 at 6:00
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    $\begingroup$ One can use lower polynomial bounds on the sine function to quickly reduce all these inequalities, for one or two variables, to polynomial inequalities, which can be then verified purely algorithmically; cf. Remark 1.7 in the note at works.bepress.com/iosif-pinelis/16/download . $\endgroup$ – Iosif Pinelis Jul 30 at 13:25

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