32. From the generalization of Cauchy's inequality, we have
(ak3+1)(ak3+1)(ak+13+1)⩾(ak2ak+1+1)3,k=1,2,⋯,n,
where an+1=a1.
Multiplying them together, we get
k=1∏n(ak3+1)3⩾k=1∏n(ak2ak+1+1)3
which simplifies to
(a13+1)(a23+1)⋯(an3+1)⩾(a12a2+1)(a22a3+1)⋯(an2a1+1)