We can use a version of the
Cauchy-Schwarz Inequality to solve this problem.
Specifically, let a1, a2, a3, ..., an and b1, b2, b3, ... bn all be positive real numbers.
Then (a1^2+a2^2+a3^2+...+an^2)*(b1^2+b2^2+b3^2+...+bn^2) >= (a1*b1+a2*b2+a3*b3+...+an*bn)^2
So for our problem let a1=sqrt(k1), a2=sqrt(k2), ..., an=sqrt(kn) and b1=sqrt(1/k1), b2=sqrt(1/k2), ..., bn=sqrt(1/kn).
Then a1^2+a2^2+a3^2+...+an^2 = k1+k2+k3+...+kn = 5n-4
and b1^2+b2^2+b3^2+...+bn^2 = 1/k1+1/k2+1/k3+...+1/kn = 1
and a1*b1+a2*b2+a3*b3+...+an*bn = 1+1+1+...+1 = n
Now just plug these in to get (5n-4) * (1) >= (n)^2. Rearrange and factor to make 0 >= (n-1)*(n-4). Then 1<=n<=4.
So now its just checking the cases of n=1, n=2, n=3, and n=4.
Previous posts have evaluated these cases finding three valid solutions: {n=1, k1=1}, {n=3, k1=2, k2=3, k3=6}, and {n4=, k1=4, k2=4, k3=4, k4=4}
Solution
