$\frac{3\sqrt7}{4}$

My idea/what I was able to think of: Using Heron's formula, x^2=s(s-a)(s-b)(s-c), so (4x)^2 is an integer. We can write this as (4x)^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c). To make x minimal we put denominator x (as in the solution above). Now it's clear that because the 4 linear factors have an even sum pairwise, they are all odd. Let w=(a+b+c), y=(-a+b+c), z=(a-b+c), t=(a+b-c). Now also use w=y+z+t and try to bound 4x to get 4x>=3*sqrt(7).