... +> ;1 2 n n 1a ,a ,...,a ,a 0. Khi đó, bằng cách đặt++=⋅ ⋅ 11 n 11 2 n 1aba a ... a,++=⋅ ⋅ ⋅22n 11 2 n 1aba a ... a,...++++=⋅ ⋅ ⋅n 1n 1n 11 2 n 1aba a ... a,ta được ( )− +⋅ ⋅ ⋅ ⋅ ⋅ =1 2 n 1 n n 1b ... +1 2 n n 1b b ... b b 1 n 1, nghóa là+ ++ ++ + +⋅ ⋅ ⋅ ⋅ ⋅ 1 2n 1 n 11 2 n 1 1 2 n 1a a...a a ... a a a ... a++ ++ ++ + ≥ +⋅ ⋅ ⋅ ⋅ ⋅ ⋅n n 1n 1 n 11 2 n 1 1 2 n 1a an 1a a ... a a a ... a 12 và bất ... 0 1 0 n 1 1 1n n nC a b C C a b ...()− + − − ++ + +n n 1 n 1 n n n n n n 1n n nC C a b C a b+ − + −+ += + + +0 n 1 0 0 1 n 1 1 1n 1 n 1C a b C a b ...( )+ − ++ − + ++ ++ +n 1 n 1n n 1 n n n 1...