文 [1]给出了条件 x+ y=1下 1xn+ λyn的最小值定理 ,并利用 (a2 + b2 ) (c2 + d2 )≥ (ac+ bd) 2 (a,b,c,d∈ (0 ,+∞ )和待定系数法证明之 .定理　已知 x,y,λ∈ (0 ,+∞ )且 x+ y=1,则当且仅当 y∶ x=λ1n+ 1 时 ,1xn+ λyn(n∈N* )取最小值 ,最小值为 (1+ λ1n+ 1 ) n+ 1 . The paper [1] gives the minimum theorem for 1xn+λyn under the condition x+ y=1 and uses (a2 + b2) (c2 + d2)≥ (ac+ bd) 2 (a,b,c,d∈ (0, +∞ ) and the undetermined coefficient method proves that the theorem knows x, y, λ∈ (0, +∞) and x+ y=1, then if and only if y: x=λ1n+ 1, 1xn+ λyn (n∈N *) Take the minimum value and the minimum value is (1+ λ1n+ 1) n+ 1.