Suppose we want to solve the following non-convex optimization problem
$$\underset{\{x_n\}} {\mathrm{argmin}} \;\;\frac{\sum_{n=1}^{N}x_n^3}{(\sum_{n=1}^{N}x_n^3)(\sum_{n=1}^{N}x_n^5)-(\sum_{n=1}^{N}x_n^4)^2}$$ $$\text{subject to} \sum_{n=1}^{N}x_n=1, \;\;\;0\leq x_n \leq1$$
It is clear that for $N=1$, the objective function is $\infty$
Also, it is clear that $\{x_n\}$ should have at least two different values. If they are equal, the objective function will be $\infty$
Let the objective function be $W$, I managed to solve the problem for $N=2$ and $N=3$ by checking the following points:
- critical points (which satisfy $\frac{\delta W}{\delta x_n}=0 \;\; \forall n$)
- corner points (for example $x_3=0$ and $x_1\neq x_2\neq0)$
I found this optimal solution for these cases: $$x_1=0.71118, \;\;\;x_2=0.28882\;\;\;\;\;\text{for}\;\; N=2$$ $$x_1=0.71118, \;\;\;x_2=0.28882, \;\;\;x_3=0 \;\;\;\;\;\text{for}\;\; N=3$$
By increasing $N$ and solving the problem, I found this interesting solution:
$$\text{For any $N\geq2$;}\;\;\;\;x_1=0.71118, \;\;\;x_2=0.28882, \;\;\;x_n=0\;\;\;\forall n\geq3$$
My question is that: Is there a method to prove the above general solution for any $N\geq2$
In other words, How to prove that increasing $N$ to be more than 2 does not decrease the objective function?