I see no connection - 2

$\begin{aligned} \displaystyle \frac{a^3+b^3+c^3}{3}-abc &= \frac{1}{3}(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\end{aligned}$

Without loss of generality, let $a>b>c$ . Then $a-b\geq 1$ , $b-c\geq 1$ , $a-c\geq 2$ . It follows that

$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2] \geq 3$

Now, the equation becomes

$\begin{aligned} \displaystyle \frac{a^3+b^3+c^3}{3}-abc &= \frac{1}{3}(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \\ & \geq a+b+c \end{aligned}$

$\begin{aligned} (a+b+c)^2 &= a^2+b^2+c^2+2ab+2bc+3ca \\ &= a^2+b^2+c^2-ab-bc-ca+3(ab+bc+ca) \\ & \geq 3 + 3(299) \\ &\geq 900 \\ a+b+c &\geq 30\end{aligned}$

Here, we used the assumption that the minimum value of $ab+bc+ca=299$ can be achieved. To show that, we must have

$\begin{aligned} ab+bc+ca = 3b^2-1 = 299 \end{aligned}$

which can be reached when $b = 10$ .

Substitute into the previous equation we get

$\begin{aligned} \displaystyle \frac{a^3+b^3+c^3}{3}-abc &= \frac{1}{3}(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \\ & \geq a+b+c \\ &\geq 30\end{aligned}$

Copy-pasted and edited from the solution for "I see no connection":

First, we notice that $\dfrac{a^3+b^3+c^3}{3}-abc=\dfrac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{3}$

In order to minimize this expression, let's first see what we must do to minimize $a^2+b^2+c^2-ab-bc-ca$ (it's clear that to minimize $a+b+c$ we want $a,b,c$ as small as possible).

Now, WLOG let $a < b < c$ . This means $a=b-m$ and $c=b+n$ .

Plugging these substitutions in the second part that we wanted to minimize, we see that $a^2+b^2+c^2-ab-bc-ca=m^2+mn+n^2$ which is minimized when $m,n$ are minimized. Thus, $m=n=1$ .

Now we know that the three variables are in fact $b-1,b,b+1$ , so it remains to minimize $b-1+b+b+1=3b$ . But this is done by minimizing $b$ .

Subbing our values of $a,b,c$ in the condition gives $b(b-1)+b(b+1)+(b-1)(b+1)=3b^2-1\ge 299\implies b\ge 10$

Thus, our expression attains minimum at $(a,b,c)=(9,10,11)$ which gives a minimum of $\dfrac{9^3+10^3+11^3}{3}-(9)(10)(11)=\boxed{30}$

If were allowed a,b and c equal the minimum value trivially will be 0 and all of them equal to 10, since the maximum volume of a parallelepiped is a cube.

So we should seek our solution with an unit difference between integers a,b ,c, .and closest to 10 Just at simple sight tell us the sum of product (10-1) 10 (10+1)=299

$a^3+b^3+c^3 - 3abc=(a+b+c)\{(a^2+b^2+c^2 - (ab+bc+ca)\}\\\text{To minimize the second bracket on the right should be minimize. }\\ \implies~ a^2+b^2+c^2~\text{ must be as near 299 as possible and greater than 299.} \\ 3*x^2 \ge 299.\text{ For x to be integer x=10. But a, b, c has to be distinct.}\\ \text{ So we can try a=9, b=10, c=11. Substituting we get needed answer as 30.}\\ \text{ It is not possible to get any where} \text{ near 3 for the second bracket of the right. }\\ \text{So 30 is the minimum.}$