作业帮a²x+b²(1-x)≥[ax+b(1-x)]²(a≠b)
来源：学生作业帮助网 编辑：作业帮 时间：2024/07/19 20:05:27
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a²x+b²(1-x)≥[ax+b(1-x)]²(a≠b)
a²x+b²(1-x)≥[ax+b(1-x)]²(a≠b)
a²x+b²(1-x)≥[ax+b(1-x)]²(a≠b)
a²x+ b²- b² x≥a² x²+ b²( x²-2x+1)+2ab(x- x²)
整理得,
（a-b）²x≥（a-b）² x²
消去（a-b）²,
0
两边展开得(a^2-b^2)x+b^2>=(a-b)^2*x^2+2(a-b)bx+b^2, 移项化简可得(a-b)^2*x>=(a-b)^2*x^2. 由a≠b可知x>=x^2或x(x-1)<=0. 解得0<=x<=1.