求证：对任意自然数n,(1+1)(1+1/4)...(1+1/(3n-2))>三次根号下（3n+1）

设x=(1+1)(1+1/4)...(1+1/(3n-2),
y=(1+1/2)(1+1/5)...(1+1/(3n-1)),
z=(1+1/3)(1+1/6)...(1+1/(3n).
易见x>y>z.
xyz=2*3/2*4/3*5/4*6/5*7/6* ……*(3n-1)/(3n-2)*(3n)/(3n-1)*(3n+1)/(3n)
=3n+1.
所以x^3>3n+1.
从而获证.