数学卷14：设x,y∈R,且xy≠0,则[x²+（1/y²)]×[(1/x²)+4y²]的最小值为（ ）

Question

数学卷14：设x,y∈R,且xy≠0,则[x²+（1/y²)]×[(1/x²)+4y²]的最小值为（ ）

Answer 1

设x,y∈R,且xy≠0,则[x²+(1/y²)]×[(1/x²)+4y²]的最小值为（ 9）
z=[x²+(1/y²)]×[(1/x²)+4y²]
=[(x²y²+1)/y²][(1+4x²y²)/x²]
=(x²y²+1)(1+4x²y²)/(x²y²)
=(x²y²+1+4x⁴y⁴+4x²y²)/(x²y²)
=1+1/(x²y²)+4x²y²+4
=1/(x²y²)+4x²y²+5≧(2√4)+5=4+5=9
即当1/(x²y²)=4x²y²,也就是x⁴y⁴=1/4时z获得最小值9.