If we are given $3$ positive integers $a,b,c$ such that $a>b>c$ , and $91b>92c>90a$ . What is the minimal value of $a+b+c$? If we are given $3$ positive integers $a,b,c$ such that $a>b>c$ , and $91b>92c>90a$ . What is the minimal value of $a+b+c$?
I am getting the bounds of the fractions $\frac{a}{b},\frac{b}{c},\frac{c}{a}$..
But I dont know what to do next

This follows up on my former comment and proves $413=139+138+136$ is the minimal sum.
$92 c \gt 90 a \implies 2c > 90(a-c) \ge 90 \cdot 2 = 180$ therefore $c \gt 90 \iff c \ge 91$.
$91 b \gt 92 c \implies 91(b-c) \gt c \ge 91$ therefore $b-c \gt \frac{91}{91} = 1 \iff b-c \ge 2 \iff b \ge c+ 2$.
Since $b \ge c+2$ and $a \gt b$ it follows that $a-c \ge 3$ and the first inequality $2c \gt 90(a-c) \ge 270$ gives the stronger bound $c \gt 135 \iff c \ge 136$.
Using the lowest values allowed by $c \ge 136$ and $b \ge c+2$ gives $c=136$, $b=c+2=138$, and $a=b+1=139$. The triplet satisfies $91 \cdot 138 = 12558 \gt 92 \cdot 136 = 12512 \gt 90 \cdot 139 = 12510$, so the minimal sum is $139+138+136 = 413$.
