If the hexagon is ABCDEF, then either AB+CD+EF or BC+DE+FA is at most 1; WLOG the former. Then extend BC, DE, FA, to form three equilateral triangles on sides AB, CD, EF. If the value of AB+CD+EF was x<=1, then the perimeter of the triangle formed by the extensions of BC, DE, FA is precisely (2-x)+2x=2+x<=3

let's name the 6 sides of the hexagon $a,b,c,d,e,f$. let's make an equilateral triangle that is made by extending the sides of $a,c,e$ and $b,d,f$. the perimeter of the two equilateral triangles are $a+c+e+2(b+d+f)$ or $2(a+c+e)+b+d+f$. Because $a+b+c+d+e+f=2$, one of $a+c+e$ and $b+d+f$ are smaller than $1$ (it can be both $1$ too) So, there exists an equilateral triangle which's perimeter is at most $3$.