Prove $a^2 + b^2 + c^2 + ab + bc +ca \ge 6$ given $a+b+c = 3$ for $a,b,c$ non-negative real.

Question

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).

Answer 1

By AM-GM $$\sum_{cyc}(a^2+ab)-6=\sum_{cyc}(a^2+ab)-\frac{2}{3}(a+b+c)^2=$$ $$=\sum_{cyc}\left(a^2+ab-\frac{2}{3}(a^2+2ab)\right)=\frac{1}{3}\sum_{cyc}(a^2-ab)=$$ $$=\frac{1}{6}\sum_{cyc}(2a^2-2ab)=\frac{1}{6}\sum_{cyc}(a^2+b^2-2ab)\geq\frac{1}{6}\sum_{cyc}(2ab-2ab)=0.$$

Answer 2

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

Answer 3

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.

Answer 4

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.

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(I wanted to provide a summary of the third approach, but I guess I'll avoid it since it's technical)

Answer 5

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.

Answer 6

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.