Introduce coordinates as usual, so that the cube consists of all points $(x,y,z)$ with $0 \le x \le 1$, $0 \le y \le 1$, and $0 \le z \le 1$.
Instead of trying to maximize the minimum length, let us try to maximize the product of the lengths of the sides, or better, the product of the squares of the lengths. Let the vertices be $(x_i,y_i,z_i)$, with $i=1$, $2$, and $3$.
Then one of the squares of a side is $S_3=(x_1-x_2)^2 +(y_1-y_2)^2+(z_1-z_2)^2,$ and the others are equally easy to write down. By the Arithmetic Mean/Geometric Mean Inequality, we have $(S_1S_2S_3)^{1/3} \le \frac{S_1+S_2+S_3}{3}.$
Note that $S_1+S_2+S_3$ consists of three terms, one of which is $(x_1-x_2)^2+(x_2-x_3)^2+(x_3-x_1)^2$ with the other terms having the same shape, but with $x$ replaced by $y$ and by $z$.
Expand. We want to maximize $2x_1^2+2x_2^2 +2x_3^2 -2x_1x_2-2x_2x_3-2x_3x_1$ under the constraints $0 \le x_i \le 1$.
It is easy to see, using the calculus or otherwise, that the maximum is $2$, and therefore the maximum value of $(S_1+S_2+S_3)/3$ is $2$.
Thus the geometric mean of the product of the squares of the sides is $\le 2$, so the product of the lengths of the sides must be $\le 2\sqrt{2}$. It follows that the lengths of the sides cannot be all $>\sqrt{2}$. It is easy to come up with an example where we get equality.
Comment: Disappointing! I am not altogether happy to see what looked like an interesting geometric problem fall to routine work with inequalities. But a purely geometric proof seems difficult unless one does a certain amount of handwaving. And as a bonus we have obtained a somewhat stronger result.
It could be interesting to look for geometric questions that yield to the same technique (answers looking for problems). The inequalities used here are of a general character, so we ought to be able to generate and answer similar questions about hypercubes. The fact that we are dealing with triangles is not really important, as long as we work with all mutual distances in a set of points.