求y=(x2+x+1)/(x2+2x+1)(x>0)的最小值

求y=(x2+x+1)/(x2+2x+1)(x>0)的最小值
求y=(x2+x+1)/(x2+2x+1)(x>0)的最小值

求y=(x2+x+1)/(x2+2x+1)(x>0)的最小值
y=(x^2+x+1)/(x^2+2x+1)
1/y=(x^2+2x+1)/(x^2+x+1)
=(x^2+x+1+x)/(x^2+x+1)
=1+(x/(x^2+x+1))
=1+1/(x+(1/x)+1)
∵x>0
∴x+(1/x)≥2√(x×(1/x))=2
∴当且公当x=1/x,即x=1时,y有最小值
y=1/(1+1/(x+(1/x)+1))
=1/(1+1/(2+1))
=3/4

＝1－（x/x²+2x+1）=1-(1/(x+1/x+2))
往下会用不等式了吧。。。。。。

y=(x^2+x+1)/(x^2+2x+1)
=1-[x/(x^2+2x+1)]
=1-[1/(x+2+1/x)]
>=1-[1/2+2√(x*1/x)]
=1-1/4
=3/4
所以当x=1时,y最小为3/4

3/4