<s>[INST] Inequality in three variables $a,b,c$ given $abc\geq 1$ 
Let $a,b,c$ be three real positive numbers such that $abc\geq 1$. Prove that $$\frac{a}{b} + \frac{b}{c} +\frac{c}{a} ≥ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$

Here's what I have done. The expression is equivalent to 
$$a^2c+b^2a+c^2b \geq  ab+bc+ac .$$ 
Denote $f(a,b,c)=a^2c+b^2a+c^2b-(ab+bc+ac)$. We need to show that $f(a,b,c)\geq 0$. 
By AM-GM, $a^2c+b^2a+c^2b \geq 3$ (Given, $abc\geq 1$). By AM-HM, $(a+b+c)(1/a +1/b +1/c)\geq 9$ by AM-GM, min value of $a+b+c=3$ so $1/a + 1/b + 1/c \geq 9/3=3$. Thus, The minimal value of $f(a,b,c)$ is equal to the min value of $a^2c+b^2a+c^2b$ minus the min value of $ab+bc+ca=3-3=0$.
Please let me know if there is fault in argument.
 [/INST] By AM-GM we obtain:
$$\sum_{cyc}\frac{a}{b}=\frac{1}{3}\sum_{cyc}\left(\frac{a}{b}+\frac{2b}{c}\right)\geq\sum_{cyc}\sqrt[3]{\frac{ab}{c^2}}=\sum_{cyc}\sqrt[3]{\frac{abc}{c^3}}\geq\sum_{cyc}\frac{1}{a}.$$
Done!
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