例14设f(x)在(a,b)内有二阶导数,f"(x)>0 试证:对λ∈(0,1,Vx1,x2∈(an,b),有 fx1+(1-x)x2lsf(x1)+(1-)f(x2) 证不妨设x1<x2记x0=Ax1+(1-4)x2 则x,< X< 0 →x-X1=(1-)(x2-x1),x2-x0=(x2-x1) 由 Lagrange定理,有 f(x0)-f(x1)=∫(51)(1-1)(x2-x1)(1) ∫(x2)-f(x)=f(2)(x2-x1)(2) (x1<51<x<52<x2)例14 [ (1 ) ] ( ) (1 ) ( ) (0,1), , ( , ) ( ) ( , ) ( ) 0 1 2 1 2 1 2 f x x f x f x x x a b f x a b f x + − + − 试证:对 ,有 设 在 内有二阶导数, 证 不妨设 x1 x2 0 1 2 记 x = x + (1− )x 则 x1 x0 x2 (1 )( ) , ( ) x0 − x1 = − x2 − x1 x2 − x0 = x2 − x1 由Lagrange定理,有 ( ) ( ) ( )(1 )( ) (1) 0 1 1 x2 x1 f x − f x = f − − ( ) ( ) ( ) ( ) (2) 2 0 2 x2 x1 f x − f x = f − ( ) x1 1 x0 2 x2