当x>1时,不等式mx^2+mx+1≥x恒成立,则实数m的取值范围是

答：
x>1时：
mx^2+mx+1>=x
m(x^2+x)>=x-1>0
m>=(x-1)/(x^2+x)
=(x-1)*[1/x-1/(x+1)]
=(x-1)/x-(x-1)/(x+1)
=1-1/x-(x+1-2)/(x+1)
=1-1/x-1+2/(x+1)
=2/(x+1)-1/x
设f(x)=2/(x+1)-1/x
求导：f'(x)=-2/(x+1)^2+1/x^2
解f'(x)=-2/(x+1)^2+1/x^2=0
得：x^2+2x+1=2x^2
x^2-2x+1=2
(x-1)^2=2
x-1=√2
x=1+√2
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