本文用初等方法证明均值不等式的拓广,并用它解决一些函数的最值问题. 定理 设x,y∈R+,a,b为正有理数,则 aabb(x+y)a+b≥(a+b)a+bxayb ①当且仅当x/a=y/b时,①式取等号. 证明 (1)当a,b为正整数时,由算术——几何平均值不等式,有 (x+y)a+b This paper uses the elementary method to prove the extension of the mean inequality, and use it to solve the problem of the most value of some functions. Theorem Let x, y ∈ R+, a, b be positive rational numbers, then aabb (x + y) a + b ≥ (a + b) a+bxayb 1 If and only if x/a = y/b, 1 takes the equal sign. Proof (1) When a, b are positive integers, from arithmetic - geometric mean inequality, there (x +y)a+b