Let a,b,c be positive real numbers numbers such that $ a^2 + b^2 + c^2 = 3 $ Let $a,b,c\in\mathbb{R^+}$ such that $  a^2 + b^2 + c^2 = 3 $. Prove that 
$$ 
(a+b+c)(a/b + b/c + c/a) \geq 9. 
$$
My Attempt 
I tried AM-GM on the symmetric expression so the $a+b+c \geq 3$, but I found $a+b+c \leq 3$.

From Cauchy-Schwarz Inequality: $$\sum\limits_{cyc} \frac{a}{b}\ge \frac{(a+b+c)^2}{ab+bc+ca}$$
Hence, it suffices to prove: $\displaystyle \frac{(a+b+c)^3}{ab+bc+ca} \ge 9$
Squaring both sides and using $a^2+b^2+c^2 = 3$,
$$\begin{align} & (a+b+c)^6 \ge 81(ab+bc+ca)^2 = 27(a^2+b^2+c^2)(ab+bc+ca)^2 \end{align}$$
which follows from the Am-Gm Inequality with $3$ terms, $$a^2+b^2+c^2,ab+bc+ca,ab+bc+ca$$
as, 
$\displaystyle \frac{(a+b+c)^6}{27} = \left(\frac{a^2+b^2+c^2+2(ab+bc+ca)}{3}\right)^3 \ge (a^2+b^2+c^2)(ab+bc+ca)^2$
