AM-GM Inequality Involving Squares and Proof

Question

Prove:

$$(a^2 + b^2 + c^2)/3 \geq ((a + b + c)/3)^2$$ OR $$(a^2 + b^2 + c^2)/3 \leq ((a + b + c)/3)^2$$

for all $a, b, c \geq 0.$ The problem wants me to find which inequality is correct and then provide a proof for it.

I'm pretty sure the first inequality is correct (I assumed this after substituting random values as $a, b,$ and $c$). I've tried factoring the inequalities and this is what I ended up with:

$$9(a^2 + b^2 + c^2) \geq 3(a^2 + b^2 + c^2 - 2ab + 2ac + 2bc)$$

However, I don't know where to go from here. I'm also meant to be applying the AM-GM theorem to solve this problem but I'm unsure where and how to apply it in this situation. Any help would be extremely appreciated :)

Answer 1

Pedestrian:

$a,b,c \ge 0.$

The first inequality is equivalent to (Dr. Graubner):

$a^2+b^2+c^2 \ge ab+bc +ac$.

AM-GM:

$a^2+b^2 \ge 2ab$; $a^2+c^2 \ge 2ac$; $b^2+c^2\ge 2bc$;

Adding LHS and RHS of these inequalities:

$2(a^2+b^2+c^2) \ge 2(ab+ac+bc),$

and we are done.

Answer 2

Your first inequality is equivalent to $$a^2+b^2+c^2\geq ab+bc+ca$$ and this is $$(a-b)^2+(b-c)^2+(c-a)^2\geq 0$$ which is true.

Answer 3

Yes, we can use AM-GM here: $$\frac{a^2+b^2+c^2}{3}-\left(\frac{a+b+c}{3}\right)^2=\frac{1}{9}\sum_{cyc}(2a^2-2ab)=$$ $$=\frac{1}{9}\sum_{cyc}(a^2+b^2-2ab)\geq\frac{1}{9}\sum_{cyc}(2\sqrt{a^2b^2}-2ab)=\frac{2}{9}\sum_{cyc}(|ab|-ab)\geq0.$$ Also, we can use AM-GM for three variables.

Indeed, $$a^3+b^3+c^3-3abc=(a+b+c)\sum_{cyc}(a^2-ab)=$$ $$=\frac{9}{2}(a+b+c)\left(\frac{a^2+b^2+c^2}{3}-\left(\frac{a+b+c}{3}\right)^2\right)$$ and since by AM-GM $$a^3+b^3+c^3-3abc\geq0$$ and $$a+b+c\geq0,$$ we are done!