16/04/26 14:22:12.90
f(n) = (3/2)(1/n^2)-(1/n!)と置く
f(1) = 1/2 > 0
f(2) = -1/8 > 0
f(3) = 0
n ≧ 4 の時
n-2 ≧ 2
(3/2) (n-1)! ≧ (3/2) (n-1)(n-2) ≧ 3(n-1) > (n-1) +1 = n
(3/2) n! > n^2
(3/2)(1/n^2) > 1/n!
n! n f(n) = (3/2) (n-1)! - n
(n+1)! (n+1) f(n+1) = (3/2) n! - (n+1)
(n+1)! n(n+1) f(n) = (3/2) (n-1)! (n+1)^2 - n (n+1)^2
(n+1)! n(n+1) f(n+1) = (3/2) n! n - n(n+1)
(n+1)! n(n+1) {f(n) -f(n+1)} = (3/2) (n-1)! { (n+1)^2 -n(n-1)} -n(n+1){(n+1) -1}
= (3/2) (n-1)! (3n +1) -n^2 (n+1)
> {(9/2) (n-1)! -n(n+1)}n
> {4(n-1)(n-2) -n(n+1)}n
= {3(n-4)^2 +11(n-4)+4}n > 0
f(n) > f(n+1)
f(n) → 0 (n→∞) であることからも n ≧ 4 ⇒ f(n) > 0 は分かる
f(4) = 5/96
f(1) +f(2) +f(2) = 1/4 > 0よりa>1
さらにb ≧ 4 の時
f(2) +f(b) +f(c) ≦ f(2) +f(4) +f(4) = -1/48 < 0よりa>2
よって
f(a)+f(b)+f(c) = 0 ⇔ a = b = c = 3