Prove $a^2 + b^2 + c^2 + ab + bc +ca \ge 6$ given $a+b+c = 3$ for $a,b,c$ non-negative real. I want to solve this problem using only the AM-GM inequality. Can someone give me the softest possible hint? Thanks.
Useless fact: from equality we can conclude $abc \le 1$.
Attempt 1:
Adding $(ab + bc + ca)$ to both sides of inequality and using the equality leaves me to prove: $ab + bc + ca \le 3$.
Final edit: I found a easy way to prove above.
$18 = 2(a+b+c)^2 = (a^2 + b^2) + (b^2 + c^2) + (c^2 + a^2) + 4ab + 4bc + 4ca \ge 6(ab + bc + ca) \implies ab + bc + ca \le 3$
(please let me know if there is a mistake in above).
Attempt 2: multiplying both sides of inequality by $2$, we get:
$(a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$. By substituting $x = a+b, y = b+c, z = c+a$ and using $x+y+z = 6$ we will need to show:
$x^2 + y^2 + z^2 \ge 12$. This doesnt seem trivial either based on am-gm.
Edit: This becomes trivial by C-S.
$(a+b).1 + (b+c).1 + (c+a).1 = 6 \Rightarrow \sqrt{((a+b)^2 + (b+c)^2 + (c+a)^2)(1 + 1 + 1)} \ge 6 \implies (a+b)^2 + (b+c)^2 + (c+a)^2 \ge 12$
Attempt 3:
$x = 1-t-u$, $y = 1+t-u$, $z = 1 + 2u$
$(1-u-t)^2 + (1-u+t)^2 + (1+2u)^2 + (1-u-t)(1-u+t) + (1+t-u)(1+2u) + (1-t-u)(1+2u)$
$ = 2(1-u)^2 + 2t^2 + (1 + 2u)^2 + (1-u)^2 - t^2 + 2(1+2u)$
expanding we get:
$ = 3(1 + u^2 -2u) + t^2 + 1 + 4u^2 + 4u + 2 + 4u = 6 + 7u^2 + t^2 + 2u\ge 6$.
Yes, this works.. (not using am-gm or any such thing).
 A: By AM-GM $$\sum_{cyc}(a^2+ab)-6=\sum_{cyc}(a^2+ab)-\frac{2}{3}(a+b+c)^2=$$
$$=\sum_{cyc}\left(a^2+ab-\frac{2}{3}(a^2+2ab)\right)=\frac{1}{3}\sum_{cyc}(a^2-ab)=$$
$$=\frac{1}{6}\sum_{cyc}(2a^2-2ab)=\frac{1}{6}\sum_{cyc}(a^2+b^2-2ab)\geq\frac{1}{6}\sum_{cyc}(2ab-2ab)=0.$$
A: For Attempt 1, you can use the rearrangement inequality: $ab+bc+ca\le a^2+b^2+c^2$ to get:
$$ab+bc+ca\le \frac{(a+b+c)^2}{3}=3$$
For Attempt 2, you can use $a+b=3-c$:
$$(3-a)^2+(3-b)^2+(3-c)^2\ge 12 \Rightarrow \\
27+a^2+b^2+c^2-6(a+b+c)\ge 12\Rightarrow \\
a^2+b^2+c^2\ge 3,$$
which is true by QM-AM.
A: Second approach : using only AM-GM
The second approach (which is correct thus far) does work out using only AM-GM. It works using CS quite clearly, but it comes out through AM-GM as well, although you have to sum a few instances of the two variable AM-GM.
We want to prove that $x^2+y^2+z^2 \geq 12$ for $x+y+z =6$ (Note : this itself was a hint in the comments!).  Apply AM-GM to $\frac{x^2}{x^2+y^2+z^2}$ and $\frac 13$, this gives you some inequality, which I'll hide :

$\frac{x}{\sqrt{3(x^2+y^2+z^2)}}\leq \frac{x^2}{2(x^2+y^2+z^2)} + \frac 16$.

Derive similar inequalities using AM-GM for $y,z$, and add them up and rearrange to get the result.

(I wanted to provide a summary of the third approach, but I guess I'll avoid it since it's technical)
A: Try using Cauchy-Schwarz on Attempt 2. You wanted only AM-GM, so you can use this post: Prove Cauchy-Schwarz with AM-GM for three variables
A: A classic fact is that if $a+b+c=3$ then $a^{p}+b^{p}+c^{p}\ge 3$ for every $p\ge 1$ this can be proved by Hôlder, C-S when $p=2$ or Power Mean inequality.
Here suppose the converse $a^2+b^2+c^2+ab+bc+ac<6$, $(p=2)$ necessarily $ab+bc+ac<3$ adding this you get $a^2+b^2+c^2+2(ab+bc+ac)=(a+b+c)^2<9$ a contradiction.
A: By the Cauchy-Schwarz, we have
$$a^2+b^2+c^2+ab+bc+ca=\frac{(a+b)^2+(b+c)^2+(c+a)^2}{2} \geqslant \frac{[(a+b)+(b+c)+(c+a)]^2}{2 \cdot 3} $$
$$=\frac{2}{3}(a+b+c)^2 = 6.$$
Done.
