Minimum of an apparently harmless function of two variables 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

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...
 A: 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. 
