1. **Problem Statement:** Given real numbers $a, b, c, d, e$, consider the expressions: $$a(b-c),\quad b(c-d),\quad c(d-e),\quad d(e-a),\quad e(a-b)$$ We want to find the maximum number of these expressions that can be negative simultaneously. 2. **Sum of expressions:** Let's sum all five expressions: $$a(b-c) + b(c-d) + c(d-e) + d(e-a) + e(a-b)$$ Expanding each term: $$= ab - ac + bc - bd + cd - ce + de - da + ea - eb$$ Rearranging terms: $$= ab - ac + bc - bd + cd - ce + de - da + ea - eb$$ Group terms to simplify: $$= (ab - ab) + (bc - bc) + (cd - cd) + (de - de) + (ea - ea) - (ac + bd + ce + da + eb) + (ab + bc + cd + de + ea)$$ Notice that positive and negative terms cancel out in pairs, so the sum simplifies to: $$0$$ 3. **Interpretation:** Since the sum of all five expressions is zero, not all five can be negative simultaneously (because sum of negatives would be negative). 4. **Maximum negative count:** The maximum number of negative expressions is therefore at most 4. 5. **Example:** To show 4 negatives is possible, choose values such as $a=1, b=2, c=3, d=4, e=5$ and check the signs: - $a(b-c) = 1(2-3) = -1$ (negative) - $b(c-d) = 2(3-4) = -2$ (negative) - $c(d-e) = 3(4-5) = -3$ (negative) - $d(e-a) = 4(5-1) = 16$ (positive) - $e(a-b) = 5(1-2) = -5$ (negative) Here, 4 expressions are negative, confirming the maximum is 4. **Final answer:** The maximum number of negative expressions among the five is $\boxed{4}$.