question
Let's compare the numbers: $A = \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $B =\frac{b}{a} + \frac{a}{c }+\frac{c}{b}$, where $a, b, c$ are real numbers such that: $0<a<b<c$
idea
because $a<b$, $\frac{a}{b}<1<\frac{b}{a}$
because $b<c$, $\frac{b}{c}<1<\frac{c}{b}$
because $a<c$, $\frac{a}{c}<1<\frac{c}{a}$
I don't know what to do next. I tried using AM-GM but with no result. Hope one of you can help me!
Unfortunately I did not see that an answer has already been posted a few minutes before mine (I was typing out this answer when it was posted) that uses a similiar method as mine, but as beginner myself, I feel the need to explain steps more thoroughly and the thinking behind employing this method, so I am posting the following answer :
I think the approach in such questions (where we can see removing the fraction appears to simplify the question) is to first convert both sides to non - fractional terms. So as a,b,c > $0$ we can multiply A and B by abc (to remove fractional terms) and if A > B then abcA > abcB. So let $abcA = x = a^2c + b^2a + c^2b \;\&\; abcB = y = b^2c + a^2b + c^2a$.
Assume x > y (another thought process in such questions where we can't directly apply any inequalities of theorems) and subtracting x on both sides we get $y - x < 0$
Substituting x and y and clubbing square terms together we get :
$a^2(b-c) + b^2(c-a) + c^2(a-b) < 0$
Let us write c - a as (c - b) + (b - a)
So we get $a^2(b-c) + b^2((c-b)+(b-a)) + c^2(a-b) < 0 \equiv -a^2(c-b) + b^2(c-b) + b^2(b-a) - c^2(b-a) < 0$
Clubbing respective terms together we get :
$(b^2-a^2)(c-b) + (b^2-c^2)(b-a) < 0$
Using $n^2 - m^2 = (n-m)(n+m)$ we get :
$(b-a)(b+a)(c-b) + (b-c)(b+c)(b-a) < 0 \equiv (b-a)(b+a)(c-b) - (c-b)(c+b)(b-a) < 0$
Since c > b > a, (b-a) and (c-b) are positive so we can divide by (b-a)(c-b) on both sides to get :
$(b+a) - (c+b) < 0 \equiv a-c < 0$ which is true (as a < c is given)
Thus our intial assumption was true that $x > y \implies A > B$