Cho \(a < b < c < d\) và \(x = \left( {a + b} \right)\left( {c + d} \right)\), \(y = \left( {a + c} \right)\left( {b + d} \right)\), \(z = \left( {a + d} \right)\left( {b + c} \right)\). Mệnh đề nào sau đây là đúng?

Ta có:

\(x - y = \left( {a + b} \right)\left( {c + d} \right) - \left( {a + c} \right)\left( {b + d} \right)\)\( = ac + bc + ad + bd - ab - cb - ad - cd\) \( = ac + bd - ab - cd \) \(= a\left( {c - b} \right) - d\left( {c - b} \right)\) \( = \left( {a-d} \right)\left( {c-b} \right) < 0\) do \(a < b < c < d\)

Suy ra: \(x < y\).

\(x - z = \left( {a + b} \right)\left( {c + d} \right) - \left( {a + d} \right)\left( {b + c} \right)\) \( = ac + ad + bc + bd - ab - ac - bd - cd\) \( = ad + bc - ab - cd\) \( = a\left( {d - b} \right) - c\left( {d - b} \right)\) \( = \left( {a - c} \right)\left( {d - b} \right) < 0\) do \(a < b < c < d\)

Suy ra \(x < z\)

\(y - z = \left( {a + c} \right)\left( {b + d} \right) - \left( {a + d} \right)\left( {b + c} \right)\) \( = ab + bc + ad + cd - ab - bd - ac - cd\) \( = bc + ad - bd - ac \) \(= b\left( {c - d} \right) - a\left( {c - d} \right)\) \( = \left( {b - a} \right)\left( {c - d} \right) < 0\) do \(a < b < c < d\)

Suy ra \(y < z\)

Vậy \(x < y < z\)