Sum of fractions Inequality: $ {a\over b} + {b \over c} + {c \over a} \ge 3 $

Question

Prove that:

$$ {a\over b} + {b \over c} + {c \over a} \ge 3 $$

Assuming $a, b, c > 0$. I was able to prove that this is true:

$$ {a\over b} + {b \over a} \ge 2 $$

by just rearranging it to get:

$$ a^2 + b^2 > 2ab $$

I figured if I would then repeat this with a,c and then b,c; I could sum the three inequalities to get closer to the third; so I get to:

$$ {a\over b} + {b \over c} + {c \over a} + {b\over a} + {c \over b} + {a \over c} \ge 6 $$

Now because of the symmetry; it certainly seems like the top inequality should be true; but I'm having trouble arguing past this point.

Answer 1

Use the inequality between the arithmetic and the geometric mean, $$ \frac 13 \left( \frac ab+\frac bc+\frac ca \right)\ge \left( \frac ab\cdot\frac bc\cdot\frac ca \right)^{1/3} =1\ . $$

Answer 2

Just use AM-GM: $${a\over b} + {b \over c} + {c \over a} \ge 3\sqrt[3]{{a\over b}\cdot {b \over c} \cdot {c \over a}} = 3$$