文[1]利用不等式:设x1,x2∈R,y1,y2∈R+,则x21y1+x22y2≥(x1+x2)2y1+y2(1)(当且仅当x1y1=x2y2时等号成立)给出了一类二元函数最值问题的一种解题策略.受此启发,本文给出另一类二元函数最值的求法.定理设x,y∈R,a,b∈R+,则(1)当ab时,有x2a-y2b≤(x-y)2a-b(2)(2)当ab时,有x2a-y [1] uses inequality: Let x1, x2∈R, y1, y2∈R+, then x21y1+x22y2≥(x1+x2)2y1+y2(1) (If and only if x1y1=x2y2, the equal sign holds) A problem-solving strategy for the problem of the two-valued function is given. Inspired by this, this paper gives another way to find the most value of the two-variable function. Theorem Let x,y∈R,a,b∈R+, (1) When ab, there are x2a-y2b≤(xy)2a-b(2)(2) When ab, there are x2a-y