$\newcommand{\R}{\mathbb{R}}$
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.