设0<x<1, a，b属于正实数，求a^2/x+b^2/(1-x)的最小值
热心网友
a^2/x+b^2/(1-x)=a^2(x+1-x)/x+b^2(x+1-x)/(1-x)=a^2+a^2(1-x)/x+b^2+b^2x/(1-x)=a^2+b^2+a^2(1-x)/x+b^2x/(1-x),而a^2(1-x)/x+b^2x/(1-x)≥2√[a^2(1-x)/x]×[b^2x/(1-x)]=2ab,所以a^2/x+b^2/(1-x)最小值为a^2+b^2+2ab=(a+b)^2
设0<x<1, a，b属于正实数，求a^2/x+b^2/(1-x)的最小值
a^2/x+b^2/(1-x)=a^2(x+1-x)/x+b^2(x+1-x)/(1-x)=a^2+a^2(1-x)/x+b^2+b^2x/(1-x)=a^2+b^2+a^2(1-x)/x+b^2x/(1-x),而a^2(1-x)/x+b^2x/(1-x)≥2√[a^2(1-x)/x]×[b^2x/(1-x)]=2ab,所以a^2/x+b^2/(1-x)最小值为a^2+b^2+2ab=(a+b)^2