Prove $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$
I will use the following lemma (the proof below):
$$2x \geq x^2(3-x^2)\ \ \ \ \text{ for any }\ x \geq 0. \tag{$\clubsuit$}$$
Start by multiplying our inequality by two
$$2\sqrt{a} +2\sqrt{b} + 2\sqrt{c} \geq 2ab +2bc +2ca, \tag{$\spadesuit$}$$
and observe that
$$2ab + 2bc + 2ca = a(b+c) + b(c+a) + c(b+c) = a(3-a) + b(3-b) + c(3-c)$$
and thus $(\spadesuit)$ is equivalent to
$$2\sqrt{a} +2\sqrt{b} + 2\sqrt{c} \geq a(3-a) + b(3-b) + c(3-c)$$
which can be obtained by summing up three applications of $(\clubsuit)$ for $x$ equal to $\sqrt{a}$, $\sqrt{b}$ and $\sqrt{c}$ respectively:
\begin{align} 2\sqrt{a} &\geq a(3-a), \\ 2\sqrt{b} &\geq b(3-b), \\ 2\sqrt{c} &\geq c(3-c). \\ \end{align}
$$\tag*{$\square$}$$
The lemma
$$2x \geq x^2(3-x^2) \tag{$\clubsuit$}$$
is true for any $x \geq 0$ (and also any $x \leq -2$) because
$$2x - x^2(3-x^2) = (x-1)^2x(x+2)$$
is a polynomial with roots at $0$ and $-2$, a double root at $1$ and a positive coefficient at the largest degree, $x^4$.
$\hspace{60pt}$
I hope this helps ;-)
From the given inequality $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ observe that $$2(ab+bc+ac)=(a+b+c)^2-a^2-b^2-c^2$$ We can rewrite the original inequality as
$$a^2+2\sqrt{a}+ b^2+2\sqrt{b}+ c^2+2\sqrt{c}\ge9$$ since $(a+b+c)=3$. Using AM-GM
set the LHS up as follows:
$$a^2+\sqrt{a}+\sqrt{a}\ge3\sqrt[3]{a^2 \sqrt{a}\sqrt{a}}=3a$$
$$b^2+\sqrt{b}+\sqrt{b}\ge3\sqrt[3]{b^2 \sqrt{b}\sqrt{b}}=3b$$
$$c^2+\sqrt{c}+\sqrt{c}\ge3\sqrt[3]{c^2 \sqrt{c}\sqrt{c}}=3c$$
Adding the three inequalities yields
$$a^2+b^2+c^2+2(\sqrt{a}+\sqrt{b}+\sqrt{c}) \ge 3(a+b+c) =9 $$ with equality if an only if $a$=$b$=$c$=$1$.
Since $ab+ac+bc=\frac{(a+b+c)^2-a^2-b^2-c^2}{2}=\frac{9-a^2-b^2-c^2}{2}$, we need to prove that $$\sum_{cyc}\left(\sqrt{a}+\frac{a^2}{2}-\frac{3}{2}\right)\geq0$$ or $$\sum_{cyc}(a^2+2\sqrt{a}-3)\geq0$$ or $$\sum_{cyc}(\sqrt{a}-1)(a\sqrt{a}+a+\sqrt{a}+3)\geq0$$ or
$$\sum_{cyc}\left((\sqrt{a}-1)(a\sqrt{a}+a+\sqrt{a}+3)-3(a-1)\right)\geq0$$ or $$\sum_{cyc}(\sqrt{a}-1)^2\sqrt{a}(\sqrt{a}+2)\geq0.$$ Done!