Let \(a,b,c\) be positive real numbers. Find the minimum value of

\(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}.\)

Havingfun Aug 6, 2019

#1**+1 **

\(\text{Just by looking at it I maintain the answer is 3, but we'll do it formally}\\ \nabla\left(\dfrac a b + \dfrac b c + \dfrac c a\right) = 0\\ \dfrac 1 b - \dfrac{c}{a^2} = 0\\ \dfrac 1 c - \dfrac{a}{b^2} = 0\\ \dfrac 1 a - \dfrac{b}{c^2} = 0\)

\(\text{The only solution of all positive numbers is $a=b=c$}\\ \text{and the extrema value is 3}\\ \text{Suppose $c=2a$}\\ \dfrac a b + \dfrac b c + \dfrac c a = 1+\dfrac 1 2 + 2 = \dfrac 7 2 > 3\\ \text{we know 3 is either a min or a max and thus as $3 < \dfrac 7 2$ it must be a min}\)

.Rom Aug 6, 2019