均值不等式求最值!求t/t^2+t-1(t>0)的最值求t/t^2+t-1(t>0)的最值,

Question

均值不等式求最值!求t/t^2+t-1(t>0)的最值
求t/t^2+t-1(t>0)的最值,

Answer 1

t/(t^2+t-1)=1/(t+1-1/t)
分母单调递增,所以函数单调递减.
当t→0时,t/(t^2+t-1)→-1,t→+∞时,t/(t^2+t-1)→0,没有最值.