Assuming $a,b,c$ are positive reals, there is no minimum!
Consider the individual term: $\left(\frac{a + 1}{a}\right)^2 = \left(1 + \frac{1}{a}\right)^2$
For positive real $a$ this is strictly greater than $1$, but as $a \to \infty$, this can be made as close to $1$ as we want.
Thus the total sum is always strictly $\gt 3$, and can be made as close to $3$ as we want.
So there is no minimum (but there is an infimum).
If you allow $a,b,c$ to be negative, then $a=b=c=-1$ gives the minimum value of $0$, as pointed out in the comments.
Perhaps you are missing some condition? Like $a,b,c$ positive reals and $abc = 1$ or something like that?