Question

求函数y=(x^4+3x^2+3)/(x^2+1)的最小值

Answer 1

y=[(x^4+x^2)+(2x^2+2)+1]/(x^2+1)
=x^2+2+1/(x^2+1)
=(x^2+1)+1/(x^2+1)+1
≥2根号[(x^2+1)/(x^2+1)+1
=3
所以,函数y=(x^4+3x^2+3)/(x^2+1)的最小值是3

Answer 2

y=(x^4+3x^2+3)/(x^2+1)
y=[(x^2+1)^2+x^2+2]/(x^2+1)
y=x^2+1+(x^2+2)/(x^2+1)
y=x^2+1+1/(x^2+1)+1
因为X^2大于0,所以在0到正无穷大上x^2+1+1/(x^2+1)单调递增,所以当X=0时Y最小
所以Y=3

Answer 3

y=(x^4+3x^2+3)/(x^2+1)
=x^2+2+1/(x^2+1)
根据基本不等式得,最小值为3
当且仅当x=0时,成立

Answer 4

y=(x^4+3x^2+3)/(x^2+1)=(x^2+1+1)+[1/(x^2+1)]=[(x^2+1)+1/(x^2+1)]+1
大于等于[2*根号下(x^2+1)*1/(x^2+1)]+1=2+1=3
当且仅当(x^2+1)=1/(x^2+1)，即x=0时取最小值3