已知0已知0已知0可以用基本不等式来做因为x+1-x=1所以a/x+b/(1-x)=a[x+(1-x)]/x+b[x+(1-x)]/(1-x)=a+a*(1-x)/x+b*x/(1-x)+b≥a+b+2√[(a*(1-x)/x)*(b*x/(1-x))]=a+b+2√(ab)=(√a+√b)^2
已知0可以用基本不等式来做因为x+1-x=1所以a/x+b/(1-x)=a[x+(1-x)]/x+b[x+(1-x)]/(1-x)=a+a*(1-x)/x+b*x/(1-x)+b≥a+b+2√[(a*(1-x)/x)*(b*x/(1-x))]=a+b+2√(ab)=(√a+√b)^2