Question

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

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
求过程

Answer 1

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^4-2a
³+6a-1)=4/[(a-1)³+4/(1+1/a)]，当a取最大值时上式有最小值，x、y为正实数，依题意设x=sinα，y=cosα，α∈(0,π/2)，a²=1+sin2α，当sin2α=1时，a有最大值√2，则1/(x²+y)+1/(x+y²)=4/[(a-1)³+4/(1+1/a)]，(√2-1)³=5√2-7，4/(1+1/a)=8-4√2，最小值为4√2-4。

Answer 2

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

Answer 3

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