已知正实数满足x^2+y^2=1,则1/(x^2+y)+1/(x+y^2)的最小值为

显示全部楼层 · 2012-1-30 11:55:23

设x+y=a，(x+y)2=x2+y2+2xy=a2，xy=(a2-1)/2，(x+y)3=x3+y3+3xy(x+y)=a3，x3+y3=a3-3a(a2-1)/2，1/(x2+y)+1/(x+y2)=(x+y+x2+y2)/[x3+y3+xy(1+xy)]=(a+1)/[a3-3a(a2-1)/2+(a^4-1)/4]=4(a+1)/(a^4-2a3+6a-1)=4/[(a-1)3+4/(1+1/a)]，当a取最大...

千问 · 2012-1-30 11:55:23

正实数满足x^2+y^2=1
x^2=y^2 =1/2、x=y=√2 /2时，为最小值1/(x^2+y)+1/(x+y^2)=4√2- 4故：1/(x^2+y)+1/(x+y^2)的最小值为4√2- 4...

千问 · 2012-1-30 11:55:23

当x=y=√2/2 时，取最小值为 2/(1/2+√2/2)=4(√2-1)≈1.6568542494923801952067548968388...