a(n)=n,
s(n)=n(n+1)/2.
1/[s(n)]^2 = 4/[n(n+1)]^2,
f(x)= 4/[x(x+1)]^2 ,x>=1,f(x) > 0.
g(x)=ln[f(x)] = ln(4) - 2ln(x)-2ln(x+1)
g'(x) = f'(x)/f(x) = -2/x - 2/(x+1),
f'(x)=f(x)*g'(x) = 4/[x(x+1)]^2 *[-2/x - 2/(x+1)],
g''(x)=2/x^2 + 2/(x+1)^2 >0,
f''(x) = f'(x)*g'(x) + f(x)*g''(x) = f(x)*[g'(x)]^2 + f(x)*g''(x) = f(x){[g'(x)]^2 + g''(x)}>0
f(x) 为凸函数.
m>=1,n>=1时,
[f(m)+f(n)]/2 >= f[(m+n)/2] = f(p),
1/[s(n)]^2 + 1/[s(m)]^2 >= 2/[s(p)]^2.