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.

With $abc=1$, we can set :$a=\frac{y}{x}; b=\frac{z}{y}; c=\frac{x}{z}$. We have:
$\frac{x}{z}+\frac{y}{z}+\frac{z}{y}+\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\le 3+\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}$
$\Leftrightarrow 3xyz+x^3+y^3+z^3\ge xy(x+y)+yz(y+z)+zx(z+x)$.
It's true by Schur's inequality. Equality occurs if only if $a=b=c=1$
