If a, b, c are positive real numbers such that $a^2+ b^2+ c^2 = 1$ $( \frac{1}{a} +\frac{1}{b} +\frac{1}{c}) +a +b +c \geq 4\sqrt{3}$ 
If a, b, c are positive real numbers such that $a^2+ b^2+ c^2 = 1$
Show that:
$$\frac{1}{a} +\frac{1}{b} +\frac{1}{c}+a +b +c \geq 4\sqrt{3}.$$

My attempt:
First , I used Holder's : $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{27}{3(a+b+c)}$$
similarly:
$$ a+b+c \geq \frac{(a+b+c)^3}{3(a^2+b^2+c^2)} = \frac{(a+b+c)^3}{3}$$
Which gives us:
$$LHS \geq \frac{27}{3(a+b+c)}+ \frac{(a+b+c)^3}{3}$$
By AM-GM:
$$LHS \geq 2\sqrt{3} \times (a+b+c)$$
So all we need to do is prove that $$2\sqrt{3} \times (a+b+c)\geq 4\sqrt{3}$$
And  this means we need to prove $$a+b+c \geq 2 $$
Which I can't.
I am not looking for a solution , only  hints that would  guide me through solving it .
Thanks in advance for any communicated help !

By your argument, we already have
$$\frac1a+\frac1b+\frac1c+a+b+c\geq\frac9{a+b+c}+\frac{(a+b+c)^3}3=\frac9u+\frac{u^3}3\equiv f(u)$$
where $u=a+b+c>0$. Try to show that $f'(x)=0$ only has one root $x^*=\sqrt3$ when $x>0$ and $f''(x^*)>0$, therefore $f(u)\geq\min_{x>0}f(x)=f(x^*)=4\sqrt3$.
