Let us substitute $b = 51 - a - c$ into the second equation to obtain: $ac(51 - a - c) = 4000$; or $51ac - a^2c - ac^2 = 4000$; or $ac^2 + (a^2 - 51a)c + 4000$; or $c = \frac{(51a - a^2) \pm \sqrt{(a^2 - 51a)^2 - 16000a}}{2a} \ge 25$; or $\pm \sqrt{(a^2 - 51a)^2 - 16000a}{2a} \ge a^2 - a$; or $(a^2 - 51a)^2 - 16000a \ge a^4 - 2a^3 + a^2$; or $0 \ge 100a^3 - 2600a^2 + 16000a$; or $0 \ge a^3 - 26a^2 + 160a$; or $0 \ge a(a - 10)(a - 16)$; or $a \in (-\infty, 0] \cup [10, 16]$ (NOTE: $0 < a \le 10$); or a = 10. This result leaves us $b + c = 41$ and $bc = 400$, which is solved according to: $c(41 - c) = 400$; or $0 = c^2 - 41c + 400 = (c - 16)(c - 25)$; or $c = 16, 25$ (NOTE: $c \ge 25$); or c = 25 and b = 16. Thus, the only solution is $\fbox{(a, b, c) = (10, 16, 25)}$.