is it $[1,\frac{\sqrt{5}+1}{2}]$? I just find its maximun and minimum.

$\frac{b}{a}=m \; , \; \frac{c}{b}=n$ $a \leq b \leq c \; \Longrightarrow \; m,n \geq 1$. Equality holds when triangle ABC is equilateral. $a+b>c \; \Longleftrightarrow \; m+1>mn$ $i) \; m \leq n \; \Longrightarrow \; m \leq n < 1+\frac{1}{m} \; \Longrightarrow \; min\{m,n\}=m<\frac{\sqrt{5}+1}{2}$ $ii) \; n \leq m \; \Longrightarrow \; n \leq m < \frac{1}{n-1} \; \Longrightarrow \; min\{m,n\}=n< \frac{\sqrt{5}+1}{2}$ We can easily show that there are positive reals $m,n$ such that $min\{m,n\}=\frac{\sqrt{5}+1}{2}-\epsilon$ for all $0<\epsilon<\frac{\sqrt{5}-1}{2}$ and hence the answer is $\left[1,\frac{\sqrt{5}+1}{2}\right)$

hi Solution is equal to (1,(1+√5)/2) because Let a triangle in a plane with a≥b≥c arbitrary. now we fix a and c and change the value of b. so we get a continues function of b that set of value of that determine with min{a/b,b/c}.clearly this function is bounded so it have suprimum wherever a/b=b/c or equivalent b^2=ac. Hence we find the state that satisfying 1. b+c>a (triangle rule), 2. b^2=ac ,3. a≥b≥c>0 Equivalent bc+c^2>ac=b^2 Equivalent (b/c)^2-b/c-1<0 So ,(1-√5)/2<b/c=a/b<(1+√5)/2 .but by 3 we get b/c>1 and a/b>1