What? Geometry?

Anybody can try $a=b$ and get the answer as 1, but how does it happen?

By using difference of two squares on the first equation,

$64 = (a+b+4)(a+b-4)(a-b+4)(-a+b+4)$ ,

which can be divided by 16 and taking square root on both sides to get

$2 = \sqrt{\frac{a+b+4}{2} \cdot \frac{a+b-4}{2} \cdot \frac{a-b+4}{2} \cdot \frac{-a+b+4}{2}}$

Looks familiar? Well, it should be, since this is the formula for the area of a triangle with sides $a,b,4$ by Heron's formula. We can see then that the triangle has area 2.

Now, we can see that the altitude to the base with length $4$ is $1$ , since $Area = bh/2 = 2$ , giving us $2h=2, h=1$ .

Then, by Apollonius' Theorem, the length of the median $d$ to the side with length $4$ is $d = \sqrt{\frac{a^2+b^2-2(\frac{4}{2})^2}{2}}= \sqrt{\frac{a^2+b^2-8}{2}}$ . Now, this is exactly what we want to minimize.

Now, in any given triangle, the median to a side is always greater than or equal to the altitude to that side. So,

$d = \sqrt{\frac{a^2+b^2-8}{2}} \geq 1$ , since the altitude to the side of length 4 is 1.

So, our answer is $\boxed{1}$

Solution by my teaming with my friend... Start by expanding: $64 = ((a + b)^2 - 16)(16 - (a - b)^2)\\$ $64 = 16((a + b)^2 + (a - b)^2)) - [(a+b)(a-b)]^2 - 256\\$ $320 = 32(a^2 + b^2) - (a^2 - b^2)^2\\$ $320 + (a^2 - b^2)^2 = 32(a^2 + b^2)\\$ To minimize RHS means minimize LHS, which means $(a^2 - b^2)^2$ should be as small as possible. Hence we let $(a^2 - b^2)^2=0$ , consequently $(a^2 + b^2) = 10$ , substitute it into the square root yields the answer $1$