笔者最近在帮助高三同学数学答疑过程中,遇见下面一道数学求最值问题:已知x,y∈R,x2+y2-3xy=2,求x2+y2的最值.解法一利用换元法结合基本不等式求解令t=xy,u=x2+y2,代入得u-3t=2,即t=u3-2又因为(1)若x=0,得y2=2,此时u=x2+y2=2.(2)若x≠0,得y=xt,此时u=x2+y2=x2+t2x2≥2t u2≥4t2 I recently in the process of helping senior high school students in mathematics answering questions, meet the following mathematics seeking the most value problem: Known x, y∈R, x2 + y2-3xy = 2, find the most value of x2 + y2. Substituting u-3t = 2 for t = xy and u = x2 + y2, we can get t = u3-2 and (1) if x = 0, we get y2 = 2 and u = x2 + y2 = 2. (2) If x ≠ 0, y = xt, then u = x2 + y2 = x2 + t2x2 ≥ 2t u2 ≥ 4t2