Prove that $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ $\geq$ $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

Question

Let $a,b,c$ be the sides of a triangle and $a+b+c=3$

Prove that:

$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

My Attempt

I think this inequality could be proved using the A.M. - G.M. however, I am not sure if I am correct or not. I request someone to please post the solution for the problem as soon as possible.

Answer 1

It's $$\sum_{cyc}a^2c\geq\sum_{cyc}ab$$ or $$3\sum_{cyc}a^2c\geq\sum_{cyc}ab\sum_{cyc}a$$ or $$\sum_{cyc}(2a^2c-a^2b-abc)\geq0$$ or $$\sum_{cyc}(4a^2c-2a^2b-2abc)\geq0$$ or $$\sum_{cyc}(a^2b+a^2c-2abc)\geq3\sum_{cyc}(a^2b-a^2c)$$ or $$\sum_{cyc}c(a-b)^2\geq\sum_{cyc}(b-a)^3$$ or $$\sum_{cyc}(a-b)^2(a+c-b)\geq0$$ and we are done!