不妨设a0<λ<1,则0<1-λ<1因为 λx1=λx1≤λx2 (1)(1-λ)x1≤(1-λ)x2=(1-λ)x2 (2)(1)+(2):λx1+(1-λ)x1≤λx1+(1-λ)x2≤λx2+(1-λ)x2 得x1≤λx1+(1-λ)x2≤x2又a所以 a<λx1+(1-λ)x2
