I have this statement:
If $\frac{a_1 + a_2 + a_3}{3} = p$, with $\{a_1,a_2,a_3\},p \in \mathbb{Z^+}$ and $p \neq 0$
Prove this: $Max(a_1\cdot a_2 \cdot a_3) = p^3$
My attempt was:
Any term can be written in the form: $a_1 = p+a, a_2 = p + b, a_3 = p +c$
So, $p + a + p + b + p + c = 3p$, thus $a+b+c= 0$
Then, i need to prove: $(p+a)(p+b)(p+c) < p^3$
$(p^2+p(a+b)+ab)(p+c) < p^3$
$p^3 + p^2(a+b) + pab + p^2c +pc(a+b)+abc < p^3$
$p^2(a+b+c)+p(ab +c(a+b)) + abc < 0$, since $a+b+c=0$
$ p(2ab+2ac+2bc)+2abc < 0$, Multiply by $2$
$p[(a+b+c)^2-(a^2+b^2+c^2)] + 2abc < 0$
$-p \cdot(a^2+b^2+c^2) + 2abc < 0$
But I don't know what else to do. Any hint is aprecciated and also i want to know the general proof of this $n$ terms. Thanks in advance.