Minimum of an apparently harmless function of two variables 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...
Any suggestion is welcomed!
 A: Here's an approach that almost works. I haven't been able to find a nice argument to establish the inequality $(\spadesuit)$ below, but verifying the inequality using the code
Reduce[(1 - Cos[Pi*x])^2 - (2 + Cos[Pi*x])*(8 x^4 - 4 x^8) > 0 && 0 < x < 1, x]

in Mathematica gives me the result
0 < x < 1

so I am almost certain the inequality is true. (I get a warning that numeric integration is being used to verify the completeness of the solution set.) If I find an appropriate argument, I'll edit the post.

Since $f$ is symmetric in $x$ and $y$, it suffices to show that $f(x,y)\geq 2\sqrt2$ holds for all $0<x<y<1$. (For points on the boundary, the inequality follows by continuity.)
Define $\phi:(0,1)\to(0,2)$ by $$\phi(x)=1-\cos\pi x.$$ Note that this is an increasing bijective function. For all $x\in(0,1)$ we have $$\phi(x)\geq2x^2,\tag{$\clubsuit$}$$ as can be easily seen using elementary calculus.
To avoid dealing with square roots, we will analyse the square of your function: $$f(x,y)^2=\frac{x^2+y^2}{x^4y^4}\frac{\phi(x)^2\phi(y)^2}{\phi(x)(3-\phi(y))+\phi(y)(3-\phi(x))}.$$ Since $\phi$ is strictly increasing and $x<y$, we have: $$\phi(x)(3-\phi(y))+\phi(y)(3-\phi(x))\leq2\phi(y)(3-\phi(x)).$$ Therefore, $$f(x,y)^2\geq\frac{x^2+y^2}{x^4y^4}\frac{\phi(x)^2\phi(y)^2}{2\phi(y)(3-\phi(x))}\geq\frac{x^2+y^2}{x^4y^2}\frac{\phi(x)^2}{3-\phi(x)}.\tag{$\ast$}$$ where in the second step we have cancelled one of the $\phi(y)$ and used $(\clubsuit)$ to get rid of the other one.
According to Mathematica, the following inequality is true for $x\in(0,1)$: $$\frac{\phi(x)^2}{3-\phi(x)}\geq8x^4-4x^8.\tag{$\spadesuit$}$$
Using this in $(\ast)$ and cancelling $x^4$, we immediately have: $$f(x,y)^2\geq\frac{(8-4x^4)(x^2+y^2)}{y^2}.$$ It remains to show that this last expression is bounded below by $8$, or equivalently: $$(8-4x^4)(x^2+y^2)\geq 8y^2.$$ But this is true, since $$(8-4x^4)(x^2+y^2)-8y^2=4x^2 (2-x^2(x^2+y^2))\geq0$$ obviously holds for $0<x<y<1$.
A: It is enough to show that both ((x^2 + y^2)^1/2 )/ (x^2 y^2) and the other part 
have minimum at (1,1) :
For ((x^2 + y^2)^1/2 )/ (x^2 y^2) the minimum happens at (x,y)=(1,1) and the 
minimum is (2^1/2)
For the other part:
Put  1-cosx = A , a-cosx = B  
0 < A < = 2 , 0 < B < = 2
Find the minimum point for AB / ((A)(3-B)+(3-A)(B))^1/2  when  0 < A <= 2 , 0 < B < = 2
It happens at (A,B)=(2,2) which means (x,y)=(1,1). The minimum of this is 2.
So the minimum of f(x,y) happens at (x,y)=(1,1) and this is 2(2^1/2)
