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DISCLAIMER: I already posted this question on Mathematics a month ago, here. However, since it has not been solved yet on that platform, I decided to ask it also here on mathoverflow. At a first glance, it looks like a straightforward calculus exercise, but it seems to hide some intrinsic difficulty... (at least, to me! :D)


I would like to prove that the minimum of the function

$$ f(x,y):=\frac{(1-\cos(\pi x))(1-\cos (\pi y))\sqrt{x^2+y^2}}{x^2 y^2 \sqrt{(1-\cos(\pi x))(2+\cos(\pi y))+(2+\cos(\pi x))(1-\cos(\pi y))}} $$

over the domain $[0,1]^2$ is $2\sqrt{2}$. Looking at the 2D plot of the function

Plot of f

one immediately notices that the minimum is $f(1,1) = 2\sqrt{2}$. However, I can't figure out how to prove this in a rigorous way, even if the expression of $f$ seems to have a nice, "quasi-separable" structure...

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    $\begingroup$ From the picture, a reasonable conjecture seems that $f$ is decreasing in both variables. So the simpler way to get your result should be: building $f$ starting by simpler monotone functions by means of operations that preserve (or invert) the convenient monotonicity. Plotting graphs of these simpler components may help. (For instance: $(1-\cos(\pi x))/x^2$ is decreasing on $[0,1]$, and so on) $\endgroup$ Commented Sep 29, 2015 at 16:23
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    $\begingroup$ @PietroMajer, what you are suggesting is exactly what I was trying to do at the beginning... but unfortunately your conjecture is not true, since $f(x,1)$ is not decreasing in $x$. $\endgroup$
    – Paglia
    Commented Sep 30, 2015 at 8:12
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    $\begingroup$ Another strategy: prove that $f$ is concave w.r.to $x$ (hence w.r.to $y$ as well, since it's symmetric). Then its global minimum must be one of the four vertices of the square. $\endgroup$ Commented Sep 30, 2015 at 18:29
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    $\begingroup$ Also, maybe maximizing $1/f^2$ is easier $\endgroup$ Commented Oct 1, 2015 at 8:29
  • $\begingroup$ Thanks Pietro! I will try to figure out if your suggestions are able to solve the problem.. $\endgroup$
    – Paglia
    Commented Oct 1, 2015 at 9:15

2 Answers 2

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The problem can be restated as follows: show that $g\ge g(\pi,\pi)$ on the square $[0,\pi]^2$, where $$g(x,y):=\frac{s(x) s(y) \sqrt{x^2 + y^2}}{2 \sqrt{2}\, r(x,y)},\quad s(x):=\frac{1-\cos x}{x^2/2},$$ $$r(x,y):=\sqrt{3 w(x)+3 w(y)-w(x) w(y)},\quad w(x):=x^2 s(x)=2(1-\cos x)\in[0,4] $$ for $(x,y)\in[0,\pi]^2$, with $s(0):=1$.

Note that $[0,\pi]^2=\bigcup_{i,j=0}^{n-1} q_{i,j}$, where $n=27$ and $q_{i,j}:=[\frac in\,\pi,\frac{i+1}n\,\pi]\times[\frac jn\,\pi,\frac{j+1}n\,\pi]$. It is enough to show that $g\ge g(\pi,\pi)$ on each of the squares $q_{i,j}$. This is easier to do for all the squares $q_{i,j}$ with $(i,j)\ne(26,26)$, because, as it turns out, $g$ is bounded away from $g(\pi,\pi)$ on each such square. The case of $(i,j)=(26,26)$ is harder, because obviously $g$ is not bounded away from $g(\pi,\pi)$ on the square $q_{26,26}$.

However, to deal with the square $q_{26,26}$, it is enough to show that $g$ is decreasing within this square along each line parallel to the diagonal from $(0,0)$ to $(\pi,\pi)$. To see this, note that, by the symmetry $g(x,y)\equiv g(y,x)$, one has $\frac d{dt}\,g(x+t,y+t)\big|_{t=0}=g_1(x,y)+g_1(y,x)$, where $g_1(x,y):=\frac d{dt}\,g(x+t,y)\big|_{t=0}$. Next, $4 \sqrt{2} \sqrt{x^2 + y^2} r(x,y)^3[g_1(x,y)+g_1(y,x)]$ is a certain polynomial (say $P$) in $x,y,u,v,u_1,v_1$, where $u:=s(x)$, $v:=s(y)$, $u_1:=s'(x)$, $v_1:=s'(y)$. Details on this and what follows can be found in the [Mathematica notebook] or its [pdf copy]. So, concerning the square $q_{26,26}$, it is enough to show that $P\le0$.
The derivative of $P$ in $u_1$ is $v (x^2 + y^2)[3w(y)+r(x,y)^2]\ge0$, so that $P$ increases in $u_1$ and, by the symmetry, in $v_1$. So, one may replace $u_1$ and $v_1$ each by $$(1)\qquad \max_{[26\pi/27,\pi]}s'=s'(\pi),$$ thus obtaining a polynomial, say $Q$, in $x,y,u=s(x),v=s(y)$, and now we need to show that $Q\le0$ over the square $q_{26,26}$. The equality (1) follows because $s'$ is increasing on $[26\pi/27,\pi]$, which is not hard to see using the general l'Hospital-type rule for monotonicity -- see e.g. [here] or [here]. Next, one can represent $Q$ as the sum of 5 terms, say $t1,\dots,t5$, of which $t1=3 \pi^3 u (u - v) v x (x - y) y\le0$, since $u=s(x),v=s(y)$ and the function $s$ is decreasing on $[0,\pi]$ (which can be easily checked by using the special l'Hospital-type rule for monotonicity). Also, $t2=-24 (u^2 x^4 + u^2 x^2 y^2 + u v x^2 y^2 + v^2 x^2 y^2 + v^2 y^4)$ and $t5=-12 u v (x^4 + y^4)$ are obviously decreasing in each of the nonnegative arguments $x,y,u,v$, whereas $t3=\pi^3 u^2 v^2 x y (x^3 + y^3)$ and $t4=4 u v (u + v) x^2 y^2 (x^2 + y^2)$ are obviously increasing in $x,y,u,v$. Therefore, we can bound $t3$ and $t4$ from above over the square $q_{26,26}$ by replacing there $x$ and $y$ each by $\pi$, and $u=s(x)$ and $v=s(y)$ each by $s(26\pi/27)$ (recall that $s$ is decreasing). Similarly, bound $t2$ and $t5$ from above by replacing $x$ and $y$ each by $26\pi/27$, and $u=s(x)$ and $v=s(y)$ each by $s(\pi)$. Thus, we indeed see that over $q_{26,26}$ one has $Q\le0$, whence $P\le0$, whence $g$ is indeed decreasing within the square $q_{26,26}$ along each line parallel to the diagonal from $(0,0)$ to $(\pi,\pi)$.

To deal with the remaining squares $q_{i,j}$ with $(i,j)\ne(26,26)$, note that
$$8g(x,y)^2=G(x,y,u,v):=\frac{u^2 v^2 (x^2 + y^2)}{3 u x^2 + 3 v y^2 - u v x^2 y^2}, $$ where, as before, $u=s(x)$ and $v=s(y)$. The partial derivative of $G$ in $u$ is $$\frac{u v^2 (x^2 + y^2)[3w(y)+r(x,y)^2]}{r(x,y)^4}\ge0 $$ at $u=s(x)$ and $v=s(y)$. By the symmetry, the partial derivative of $G$ in $v$ is $\ge0$ as well. So, $8g(x,y)^2=G(x,y,u,v)$ will be bounded from below on each remaining square $q_{i,j}$ if we replace $u$ and $v$ by their smallest respective values, $s(\frac{i+1}n\,\pi)$ and $s(\frac {j+1}n\,\pi)$, on this square; recall that $s$ is decreasing. Using computer algebra, it is then easy to see that $8g(x,y)^2-8g(\pi,\pi)^2\ge G\big(x,y,s(\frac {i+1}n\,\pi),s(\frac{j+1}n\,\pi)\big)-8g(\pi,\pi)^2>0$ on each remaining square $q_{i,j}$. This completes the proof.

Addendum: I overlooked that one has also to show that $g(x,\pi)$ is decreasing in $x\in[26\pi/27,\pi]$. Note that $\sqrt{2} \pi^2 \sqrt{\pi^2 + x^2}\, \big(12 - x^2 s(x)\big)^{3/2}\,\frac d{dx}g(x,\pi)=T1+T2$, where $T1:=u x (24 + x^3 z_1 + \pi^2 (2 u + x z_1))$ and $T2:=-24 (\pi^2 + x^2) z_1$, where $z_1:=-u_1$ and, as before, $u=s(x)$ and $u_1=s'(x)$. So, $\frac d{dx}g(x,\pi)$ equals $T1+T2$ in sign. Recall that $s$ is decreasing and $s'$ is increasing on $[26\pi/27,\pi]$. In particular, it follows that $z_1=-s'(x)\ge-s'(\pi)=8/\pi^3>0$ for $x\in[26\pi/27,\pi]$. In the way we did before, we can now bound $T1$ from above over $[26\pi/27,\pi]$ by replacing there $x$, $u$, and $z_1$ respectively by $\pi$, $s(26\pi/27)$, and $-s'(26\pi/27)$. Also, we can bound $T2$ from above over $[26\pi/27,\pi]$ by replacing there $x$, $u$, and $z_1$ respectively by $26\pi/27$, $s(\pi)$, and $-s'(\pi)$. This will show that $T1+T2<-50<0$ over $[26\pi/27,\pi]$, and so, $g(x,\pi)$ is indeed decreasing in $x\in[26\pi/27,\pi]$.

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  • $\begingroup$ I have added an addendum on a previously overlooked point. $\endgroup$ Commented Oct 2, 2015 at 20:53
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For the function $g$ defined in my first answer and for $(x,y)\in(0,\pi]^2$ one has $$2\sqrt2\,g(x,y)=\frac{s(x) s(y)}{\sqrt{(2+\cos y)s(x)p+(2+\cos x)s(y)(1-p)}},\quad (*)$$ where $p:=\frac{x^2}{x^2+y^2}\in(0,1)$ and, as before, $s(x):=\frac{1-\cos x}{x^2/2}$. Obviously,
$$(2+\cos y)s(x)p+(2+\cos x)s(y)(1-p)\le\max[(2+\cos y)s(x),(2+\cos x)s(y)].$$ So, $$2\sqrt2\,g(x,y)\ge\min[a(x)b(y),a(y)b(x)],\qquad(**)$$ where $$a(x):=\sqrt{s(x)}, \quad b(x):=\frac{s(x)}{\sqrt{2+\cos x}}. $$ Note that the function $a$ is nonnegative and decreasing (on $(0,\pi]$), since the function $s$ is so. Clearly, the function $b$ is nonnegative. We will show that $b$ is decreasing. It will then follow by $(**)$ that for all $(x,y)\in(0,\pi]^2$ $$2\sqrt2\,g(x,y)\ge a(\pi)b(\pi)=2\sqrt2\,g(\pi,\pi), $$ as desired.

It remains to show that $b$ is decreasing. Using the substitution $x =2 \arctan t$ with $t\in(0,\infty)$ (so that $\cos x= (1 - t^2)/(1 + t^2)$), write $b(x)=B(t)^2$, where $B(t):=F(t)/\arctan t$, $F(t):=t(3 + 4 t^2 + t^4)^{-1/4}$. Note that $F(0)=\arctan0=0$. Let $\rho:=F'/\arctan'$. Then $\rho'(t)= -t^3 (1 + 2 t^2)/((1 + t^2) (3 + t^2)^2 (3 + 4 t^2 + t^4)^{1/4})<0$, so that $\rho$ is decreasing. It follows by the special l'Hospital-type rule for monotonicity (see e.g. [here]) that $B=F/\arctan$ is decreasing, and hence the function $b$ is indeed decreasing, which completes the proof.


Working a bit harder, once can show that the result will hold more generally: when (both instances of) the constant $2$ in the right-hand side expression in $(*)$ for $2\sqrt2\,g(x,y)$ are replaced by any real $c\ge c_*$, where $$c_*:=\frac{\pi^4+16}{\pi^4-16}=1.393\dots.$$ If we only assume that $c\in(0,\infty)$, we can no longer maintain that the function $b$ is decreasing. However, the inequality $b(x):=\frac{s(x)}{\sqrt{c+\cos x}}\ge b(\pi)=\frac{s(\pi)}{\sqrt{c-1}}$ that we need for $x\in(0,\pi]$ can be rewritten as $c\ge h_*:=\sup_{x\in(0,\pi]}h(x)$, where $$h(x):=\frac{s(x)^2 + s(\pi)^2 \cos x}{s(x)^2 - s(\pi)^2}. $$ Since $h(0+)=c_*$, it is clear that $h_*\ge c_*$, and so, to have $b\ge b(\pi)$ it is necessary that $c\ge c_*$.

On the other hand, let us show that the condition $c\ge c_*$ is also sufficient for $b\ge b(\pi)$. Since $b\ge b(\pi)$ has been shown to be equivalent to $c\ge h_*$, it suffices to show that $b\ge b(\pi)$ holds for $c=c_*$. Next, the inequality $b(x)\ge b(\pi)$ can also be rewritten as $$0\le\Big(\frac{4 (1 - \cos x)^2}{(c + \cos x) b(\pi)^2}\Big)^{1/4}-2x$$ $$=d(t):= \pi t\, \Big(\frac{c-1}{(1 + t^2) (1 + c - t^2 + c t^2)}\Big)^{1/4} - 2\arctan t, \qquad(***) $$ where, as before, $x =2 \arctan t$ with $t\in(0,\infty)$. Further, $d'(t)(1 + t^2)=u(t)/v(t)-2$, where $u(t):=(c - 1)^{1/4} \pi (1 + t^2) (1 + c + c t^2)$ and $v(t):=(1 - t^4 + c (1 + t^2)^2)^{5/4}$, so that $d'(t)$ is equal in sign to $d_1(t):=u(t)^4-2v(t)^4$, which is a polynomial. Therefore, one can see that, for $c=c_*$ and some $t_*\in(0,\infty)$, one has $d_1>0$ on $(0,t_*)$ and $d_1<0$ on $(t_*,\infty)$; in fact, $t_*=8.657\dots$. So, $d$ increases on $(0,t_*)$ and decreases on $(t_*,\infty)$. At that, $d(0+)=0=d(\infty-)$. So, $d>0$ on $(0,\infty)$, and the inequality in $(***)$ is proved.

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  • $\begingroup$ My second answer was an attempt to use the same basic idea, with $p$ and $1-p$, as in my latest answer, but the calculations there were a bit off. So, I am going to delete my second answer. The remaining two answers are based on very different ideas, and so, I don't think it would be helpful to combine the two answers into one. It also seems to me that the latest answer is much more elegant than the first one. $\endgroup$ Commented Oct 4, 2015 at 2:41
  • $\begingroup$ I have now provided details on the generalization, with $c\ge c_*=1.393\dots$ instead of $c=2$. $\endgroup$ Commented Oct 4, 2015 at 17:55

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