Question

函数y=4x^2+16/(x^2+1)^2的最小值

Answer 1

设x^2+1 = t那么t>=1

y=4(t-1) + 16/t^2 = 2t + 2t + 16/t^2 -4 >= 3(2t*2t*16/t^2)^1/3 - 4 = 3*8-4=20

当2t=16/t^2也就是t=2的时候取得最小值20

此时x^2 = 1