Simple AM-GM inequality Let $a,b,c$ positive real numbers such that $a+b+c=3$, using only AM-GM inequalities show that
$$
a+b+c \geq ab+bc+ca
$$
I was able to prove that
$$
\begin{align}
a^2+b^2+c^2 &=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{a^2+c^2}{2} \geq \\
&\ge  \frac{2\sqrt{a^2b^2}}{2}+\frac{2\sqrt{b^2c^2}}{2}+\frac{2\sqrt{a^2c^2}}{2} \\
&= ab+bc+ca
\end{align}
$$
but now I am stuck. I don't know how to use the fact that $a+b+c=3$ to prove the inequality. Anybody can give me a hint?
 A: $$9=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 3(ab+bc+ca)$$
A: Hint: Multiply the original inequality by $a+b+c$ on the LHS and $3$ on the RHS, expand and eliminate common terms and you will arrive at something you have proved.
A: Update: Upps, that is the same as that of @Vincent, sorry didn't see that first. However, it's a bit more explicte 
A nicer way to use your results and proceed frm there is the following. You've got that
$$ a^2+b^2+c^2 \ge ab+bc+ca $$
Now use the fact that $(a+b+c)=S=3$ and multiply each side of your original inequality by S this is
$$ (a+b+c)S \ge S(ab+bc+ca)$$
or, (where we use now that $S=3$)
$$ (a+b+c)(a+b+c) \ge 3(ab+bc+ca)$$
Then
$$ a^2+b^2+c^2 + 2(ab+bc+ca) \ge 3(ab+bc+ca)$$
and
$$ a^2+b^2+c^2  \ge ab+bc+ca$$
which is exactly that, what you've already proven on your own.
A: $9 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ac) \geq 3(ab+bc + ac)$
A: Reformulate first $a+b+c=S$ and $a=M+d \qquad b=M-d $ then 
$$S  \ge M^2-d^2 + (S-2M)2M$$
reorganize
$$ 3M^2-2SM+S  +d^2  \ge 0 $$
Now make use of the given definition that $S=3$. We get
$$3(M^2-2M+1)  +d^2  \ge 0 $$
$$3(M-1)^2 +d^2  \ge 0 $$
which is always true.   
Well, this focuses "when and how" it makes sense to introduce the condition that $S=3$. Unfortunately the step with the AM-GM-inequality is lost. But maybe you can combine your steps with this derivations? 
