First, I will prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3$, and then I will prove that $\displaystyle\frac{3 \sqrt[3]{abc}}{a+b+c}$.
1) $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3$
Let's call $\varphi(x_1,x_2,x_3)=\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_1}$. Le'ts prove that $\varphi$ is minimized when $x_1 = x_2 = x_3$: $\frac{\partial\varphi}{\partial x_i} = \frac{1}{x_{i+1}} - \frac{x_{i-1}}{x_i^2} = \frac{x_i^2 - x_{i-1}x_{i+1}}{x_i^2 x_{i+1}}$ (with $x_4 = x_1$ and $x_0 = x_3$).
So $\Delta\varphi = 0$ is equivalent to : $x_1^2 = x_2 x_3$ $x_2^2 = x_1 x_3$ $x_3^2 = x_1 x_2$
By dividing one equation by another, we get that $x_1 = x_2 = x_3$. We need to prove that $\varphi$ is convex, to show that this function is minimized when its gradient is $0$. For this, let's compute its Hessian matrix $M$:
$M = (M_{ij}) = (\frac{\partial^2\varphi}{\partial x_i\partial x_j})$ We have: $M_{ii} = \frac{2 x_{i-1}}{x_i^3}$ $M_{i(i+1)} = -\frac{1}{x_{i+1}^2}$ $M_{i(i-1)} = -\frac{1}{x_i^2}$
Which gives us:
$M = \left( \begin{array}{ccc} \frac{2 x_3}{x_1^3} && -\frac{1}{x_2^2} && -\frac{1}{x_1^2}\\ -\frac{1}{x_2^2} && \frac{2 x_1}{x_2^3} && -\frac{1}{x_3^2}\\ -\frac{1}{x_1^2} && -\frac{1}{x_3^2}&& \frac{2 x_2}{x_3^3} \end{array} \right)$
Calculating its determinant and verifying it is positive will show that $\varphi$ is convex, which achives to prove inequality 1)
2) $\displaystyle\frac{3 \sqrt[3]{abc}}{a+b+c}\geq 1$ This is equivalent to $\sqrt[3]{abc}\geq\frac{a+b+c}{3}$, or $\ln(\frac{a+b+c}{3})\geq \frac{1}{3}(\ln a +\ln b+\ln c)$, which is true because the function $\ln$ is concave.
By combining those two inequalities, we instantly solve the two questions.