问题已知a,b∈R~+,x,y∈R,且a+b=1,求证:ax~2+by~2≥(ax+by)~2.解法1作差比较简单明了ax~2+by~2-(ax+by)~2=ax~2+by~2-a~2x~2-b~2y~2-2abxy=a(1-a)x~2-2abxy+b(1-b)y~2=ab(x~2-2xy+y~2)=ab(x-y)~2≥0.解法2代换在前作差在后因为a+b=1,令T=(a+b)(ax~2+by~2)-(ax+by)~2=abx~2+aby~2-2abxy=ab(x-y)~2≥0.评析“作差法”是证明不等式的一种最基本的方法,巧用作差法是我们解决不等式证明问题的一种行之有效的途径,如果应用得恰当,能切中要害,问题 Problem is known a, b∈R ~ +, x, y∈R, and a + b = 1, verify: ax ~ 2 + by ~ 2 ≥ (ax + by) ~ 2. Solution 1 is relatively simple and clear ax ~ 2 + by ~ 2- (ax + by) ~ 2 = ax ~ 2 + by ~ 2-a ~ 2x ~ 2-b ~ 2y ~ 2-2abxy = a (1-a) x ~ 2-2abxy + b (1-b) y ~ 2 = ab (x ~ 2-2xy + y ~ 2) = ab (xy) ~ 2≥0. (a + b) (ax ~ 2 + by ~ 2) - (ax + by) ~ 2 = abx ~ 2 + aby ~ 2-2abxy = ab (xy) ~ 2≥0. Is the most basic method to prove the inequality. Using it as the bad method is an effective way to solve the inequality problem. If it is applied properly,