Solution:
At least two of $a, b, c$ are either not less than $1$ or not greater than $1$. Assume that $a$ and $b$ are on the same side of $1$. Next, transform the inequality as follows:
$$\begin{aligned} a(a-1)+b(b-1)+c(c-1) &\stackrel{?}{\geq} 0 \\ a(a-1)+b(b-1)+c^{2}\left(1-\frac{1}{c}\right) &\stackrel{?}{\geq} 0 \\ a(a-1)+b(b-1)+c^{2}(1-a b) &\stackrel{?}{\geq} 0 \\ a(a-1)+b(b-1)-c^{2}(a-1)-c^{2}(a b-a) &\stackrel{?}{\geq} 0 \\ \left(a-c^{2}\right)(a-1)+\left(b-c^{2} a\right)(b-1) &\stackrel{?}{\geq} 0 \\ \left(a-\frac{1}{a^{2} b^{2}}\right)(a-1)+\left(b-\frac{1}{a^{2} b}\right)(b-1) &\stackrel{?}{\geq} 0 \end{aligned}$$
Using the hypotheses concerning $a$ and $b$, it is not hard to see that the four factors in parentheses are all nonnegative or all nonpositive, and therefore the left side is nonnegative.