Prove $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$ Let $a,b,c$ are non-negative numbers, such that $a+b+c = 3$.
Prove that $\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$

Here's my idea:
$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge ab + bc + ca$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(ab + bc + ca)$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - 2(ab + bc + ca) \ge 0$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) - ((a+b+c)^2 - (a^2 + b^2 + c^2) \ge 0$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) - (a+b+c)^2 \ge 0$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$
And I'm stuck here.
I need to prove that: 
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge (a+b+c)^2$ or
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a+b+c)$, because $a+b+c = 3$
In the first case using Cauchy-Schwarz Inequality I prove that:
$(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$
$3(a^2 + b^2 + c^2) \ge (a+b+c)^2$
Now I need to prove that: 
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3(a^2 + b^2 + c^2)$ 
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a^2 + b^2 + c^2)$ 
$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge a^2 + b^2 + c^2$ 
I need I don't know how to continue.
In the second case I tried proving:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$ and
$a^2 + b^2 + c^2 \ge a+b+c$
Using Cauchy-Schwarz Inequality I proved:
$(a^2 + b^2 + c^2)(1+1+1) \ge (a+b+c)^2$
$(a^2 + b^2 + c^2)(a+b+c) \ge (a+b+c)^2$
$a^2 + b^2 + c^2 \ge a+b+c$
But I can't find a way to prove that $2(\sqrt{a} + \sqrt{b} + \sqrt{c}) \ge 2(a+b+c)$
So please help me with this problem.
P.S 
My initial idea, which is proving:
$2(\sqrt{a} + \sqrt{b} + \sqrt{c}) + (a^2 + b^2 + c^2) \ge 3^2 = 9$
maybe isn't the right way to prove this inequality.
 A: 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 $a$=$b$=$c$=$1$.
A: 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!
A: 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 ;-)
A: Hint: 

What lower bound does AM-GM give you when you consider $a^2 + \sqrt{a} + \sqrt{a}$?

Your hope that $\sum \sqrt{a} \ge \sum a = 3$ is false, by using Cauchy Schwarz: $9 = 3(\sum a) \ge (\sum \sqrt{a})^2$. In fact, when $a+b+c = 3$, we have
$$\sum a^2 \ge \sum a = 3 \ge \sum \sqrt{a}$$
all by Cauchy-Schwarz, so your hope to split the inequality up is thwarted. This also signals us that we should try to "mix" $a^2$ and $\sqrt{a}$ together in some way, hence the hint.
