The idea is that q is greater than zero from the known inequality, and the value of q is controlled by arithmetic progression.

If q is small, the number of arithmetic progression is also small.

A2 & gt=a 1, so the minimum a2 is 1, so A4 = 2 and A6 = 3.

And a3 = q, a5 = q * q, a7 = q * q * q

A 1 Existence inequality

1 & lt； = 1 & lt； = q< = 2< = q * q< = 3< = q * q, it is concluded that the cube root of 3 is the smallest real number that meets the inequality.

My writing process is not so rigorous, but that's what I do. I hope I helped you.