设x,y是正实数,且x+y=1,则x2/x+2 +y2/y+1的最小值
问题描述:
设x,y是正实数,且x+y=1,则x2/x+2 +y2/y+1的最小值
答
x²/(x+2)+y²/(y+1)=(x²-4+4)/(x+2)+(y²-1+1/(y+1)=x-2+y-1+4/(x+2)+1/(y+1) 把x+y=1带入有=4/(x+2)+1/(2-x)-2=(10-3x)/(4-x²)-2设h(x)=(10-3x)/(4-x²),对函数求导得:h '(x)=(-3x...