How do I prove this inequality?: $a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ where $a,b,c>0$ and $abc=1.$ I have been thinking over this problem for a couple of days, but I have no idea how to solve it in a simple way. 
I am interested if there is a way only using elementary methods to prove it.
Using the software Mathematica confirmed this inequality is correct.

Since $abc = 1$, any product term such as $ab$ or $\frac{1}{bc}$ can be rewritten as a singleton term  (i.e., $ab = \frac{1}{c}$ and $\frac{1}{bc} =a$.  Thus the inequality is equivalent to $$\left(a-1+\frac{1}{c}\right)\left(b-1+\frac{1}{a}\right)\left(c-1+\frac{1}{b}\right) \leq 1.$$
This inequality was Problem 2 on the 2000 International Mathematical Olympiad.  Knowing that, solutions should be easy to find.

For example, here's a nice one due to Robin Chapman that can be found here.

We have $$\left(b - 1 + \frac{1}{a}\right) = b\left(1 - \frac{1}{b} + \frac{1}{ab}\right) = b\left(1 + c - \frac{1}{b}\right).$$ Hence, $$\left(c - 1 + \frac{1}{b}\right)\left(b - 1 + \frac{1}{a}\right) = b\left(c^2 - \left(1 - \frac{1}{b}\right)^2\right) \leq b c^2.$$ 
Thus $$\left(a-1+\frac{1}{c}\right)^2\left(b-1+\frac{1}{a}\right)^2\left(c-1+\frac{1}{b}\right)^2 \leq b c^2 a b^2  c a^2 = 1,$$
proving the inequality in the case that every factor on the left side of the inequality is positive.
Now, suppose one of the factors on the left-hand side is negative; i.e., $a - 1 + \frac{1}{c} < 0$.  Then $a < 1$ and $c > 1$.  Thus $b-1+\frac{1}{a} > 0$ and $c-1+\frac{1}{b} > 0$.  So the left-hand side of the inequality is negative, and the inequality still holds.
