2010/042) Prove that if a < b and c < d then ad + bc < ac + bd

a < b

so (b-a) > 0

also (d-c) > 0
as both are positive hence

(b-a)(d-c) > 0

or bd - bc -ad + ac > 0

or (bd+ac) - (bc+ad) > 0

or (bd + ac) > (bc + ad) or (bc+ad) < (bd+ac) or ad + bc < ac + bd