Is this AM-GM application correct? I have got an inequality down to proving that: if $a,b,c$ are positive reals that satisfy $a+b+c=1$, then $$\frac{1}{1-\sqrt{a}}+\frac{1}{1-\sqrt{b}}+\frac{1}{1-\sqrt{c}}\ge \frac{9+3\sqrt{3}}{2}$$
Here is my "proof":
Since $a,b,c$ are positive reals and $a+b+c=1$, then $0<a,b,c<1$. Thus $\frac{1}{1-\sqrt{a}}$ etc are all positive, so we can use AM-GM to get $$\frac{1}{1-\sqrt{a}}+\frac{1}{1-\sqrt{b}}+\frac{1}{1-\sqrt{c}}\ge 3\sqrt[3]{\frac{1}{(1-\sqrt{a})(1-\sqrt{b})(1-\sqrt{c})}}$$
Equality occurs if and only if $\frac{1}{1-\sqrt{a}}=\frac{1}{1-\sqrt{b}}=\frac{1}{1-\sqrt{c}}$, i.e. $a=b=c$. Since $a+b+c=1$ it follows that equality occurs when $a=b=c=1/3$. Therefore (from our AM-GM application), the minimum value of $\frac{1}{1-\sqrt{a}}+...$ is $3\left(\frac{1}{1-\sqrt{1/3}}\right)=\frac{9+3\sqrt{3}}{2}$. So it follows that $$\frac{1}{1-\sqrt{a}}+\frac{1}{1-\sqrt{b}}+\frac{1}{1-\sqrt{c}}\ge \frac{9+3\sqrt{3}}{2}$$
as required
I'm not sure if this is correct, and would really appreciate if someone could check it for me.
Thanks for your help!
 A: In general, suppose you have $\mathrm D \subset \mathbb R^n$ and need to minimise $f: \mathrm D \mapsto \mathbb R$.  Note that any side conditions can be incorporated into our definition of $\mathrm D$. Further suppose you have found an inequality of form $f(x) \ge g(x)$, for any $x \in \mathrm D$ and equality is possible when $x=a \in \mathrm D$. Now the question is can you conclude $f(x) \ge g(a)$ for all $x \in \mathrm D$?
In general, the answer is no.  All we need is a counter example, several have been provided in comments above - e.g. take $\mathrm D = [0, 1]$ and $f(x) = x^2+1$.  We can write the inequality $f(x) \ge 2x$ with equality iff $x=1$ by AM-GM, which should have given the minimum as $2$, but clearly $f(0) =1$ is the minimum. The situation is perhaps more obvious from the graph below.

So under what conditions can we claim we have found a minimum? Clearly if we have the further condition that $\forall x \in \mathrm D, \; g(x) \ge g(a)$, then we can easily conclude $f(x) \ge g(a)$ for all $x \in \mathrm D$, and as $f(a)=g(a)$ can be achieved, this has to be the minimum.  Hence this is a sufficient condition.  Note that $g$ being constant is a special case of this condition and that it is not necessary that the equality is achieved at a unique value.  

P.S. For your specific inequality, one way would be to note that it is $\sum_{cyc} \left(\dfrac1{1-\sqrt a} -\dfrac{3+\sqrt3}2 \right) \ge 0$.  Using the constraint, we can add any multiple of $\sum_{cyc} (1-3a)$ without changing the inequality.  So it is enough to show that 
$$\frac1{1-\sqrt{a}}-\frac{3+\sqrt3}2 +\frac{3+2\sqrt3}4 (1-3a) \ge 0 \\ \iff \left(\sqrt{a}-\frac1{\sqrt3}\right)^2 \left(\sqrt{a}+\frac2{\sqrt3}-1\right) \ge 0$$
