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 $x\in[26\pi/27,\pi]$.