TITLE: Prove that $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$
QUESTION [9 upvotes]: Let $a,b,c\in{\mathbb{R^+}}$. Prove that 
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}$$
I tried to expand both, but I did not get anything useful.

REPLY [4 votes]: As Macavity says, this is the same as
$$
\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2} \right) +\left(\frac{b}a+\frac{a}c+\frac{c}b\right)\ge \left(\frac{a}b+\frac{b}c+\frac{c}a\right)+3
$$
which is the same after rearrangement as
$$
\underbrace{\left(\frac ab-1\right)^2+\left(\frac bc-1\right)^2+\left(\frac ca-1\right)^2}_{\ge0}
\ge\underbrace{6-\left(\frac ab+\frac bc+\frac ca\right)-\left(\frac ba+\frac ac+\frac cb\right)}_{\le0}
$$
since $x+\frac1x\ge2$.