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 $a$=$b$=$c$=$1$.