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 $a+b+c \leq 3$.

We'll prove that $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\geq\frac{3}{2}\sum\limits_{cyc}\left(\frac{a}{b}+\frac{a}{c}\right)$.
Let $\frac{a}{b}=x$, $\frac{b}{c}=y$ and $\frac{c}{a}=z$.
Thus, we need to prove that $2(x+y+z)^2\geq3(x+y+z+xy+xz+yz)$, 
where $x>0$, $y>0$ and $z>0$ such that $xyz=1$
or $\sum\limits_{cyc}(2x^2-3x+xy)\geq0$ or $\sum\limits_{cyc}\frac{(2x+1)(x-1)^2}{x}\geq0$, which is obvious. 
Hence, it remains to prove that $(a+b+c)^2\sum\limits_{cyc}(a^2b+a^2c)\geq18abc(a^2+b^2+c^2)$.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $9u^2(9uv^2-3w^3)\geq18w^3(9u^2-6v^2)$ or $(7u^2-4v^2)w^3\leq3u^3v^2$.
But $w^3$ gets a maximal value, when two numbers from $\{a,b,c\}$ are equal.
Since the last inequality is homogeneous, it remains to check one case only: $b=c=1$,
which gives $(a-1)^2(a^2-2a+4)\geq0$. Done!
