IMO 1967 LL BUL2

Origin: BUL

Problem

Prove that 1 3n2 + 1 2n + 1 6 \geq(n!)2/n (n is a positive integer) and that equality is possible only in the case n = 1.

Solution

(n!)2/n = ((1 \cdot 2 \cdot \cdot \cdot n)1/n)2 \leq
1+2+\cdot\cdot\cdot+n n 2 =
n+1 2 \leq1 3n2 + 1 2n + 1 6.