柯西不等式二维形式(a^2+b^2)(c^2 + d^2)≥(ac+bd)^2推广(a1·b1+a2·b2+a3·b3+...+an·bn)^2≤ (a1^2)+(a2^2)+(a3^2)+...+(an^2))((b1^2)+(b2^2)+(b3^2)+...(bn^2))故：(2a+3b+4c)*(1/2+1/3+1/4)≥(根号2a*根号1/2...