How to show the optimal condition of $f(\alpha) = \frac{R^2+G^2\sum_{i=1}^k \alpha_i^2}{2\sum_{i=1}^k \alpha_i}$

Question

Consider the following function: ($\alpha>0$)
$$f(\alpha) = \frac{R^2+G^2\sum_{i=1}^k \alpha_i^2}{2\sum_{i=1}^k \alpha_i}$$

1. It is a quadratic (in $\alpha$) over linear (in $\alpha$); therefore, it is a convex function.
2. It is a symmetric function (in $\alpha$) since any permutation in $\alpha_i$ won't change the value of this function. Ex: $\alpha_1+\alpha_2 + \alpha_3 = \alpha_2+\alpha_3 + \alpha_1$.

How to show that the minimal $f(\alpha)$ occurs when $\alpha_i$ are equal?

What I could do is the simplest case:

$$f(\alpha) = \frac{R^2+G^2(\alpha_1^2+\alpha_2^2)}{2(\alpha_1+\alpha_2)}$$

I don;t think $R$ will influence the result (just a constant), so we can simplify it as:

$$f(\alpha) = \frac{G^2(\alpha_1^2+\alpha_2^2)}{2(\alpha_1+\alpha_2)}$$
and it becomes:
$$f(\alpha) = \frac{G^2((\alpha_1+\alpha_2)^2-2\alpha_1\alpha_2)}{2(\alpha_1+\alpha_2)} = \frac{G^2}{2}(\alpha_1+\alpha_2)-G^2\frac{\alpha_1\alpha_2}{\alpha_1+\alpha_2}$$

How to show the following and the general case?

Answer 1

Let $A=(\alpha_1,\alpha_2,\dots,\alpha_k)$ be any optimal solution, and suppose that the values aren't equal.

Now consider all possible permutations of $A$: they must all obtain the same optimal value by symmetry.

Now consider the mean of those permutations. This mean will have the same value for each $\alpha_i$. The mean is a convex combination, and the function is convex, so the optimal value at this new point must be less than or equal to the value at $A$. You've just recovered an optimal solution with all equal entries.

This does not prove that an optimal solution exists. Nor does it prove that the only optimal solutions have equal entries. It just proves this: that if there is an optimal solution, there must be at least one with all equal entries. However, now that you've reduced it to a univariate problem, you shouldn't have difficulty establishing the optimal value, probably analytically.

If you wish to prove that solutions with unequal $\alpha_i$ do not exist, you'll want to show that the function is strictly convex. I'm pretty sure that's the case at least for positive $R$ and $G$.

Answer 2

If you consider the function to be minimized $$F= \frac{R^2+G^2\sum_{i=1}^k \alpha_i^2}{2\sum_{i=1}^k \alpha_i}$$ consider its derivatives. They write $$\frac{dF}{d\alpha_j}=\frac{G^2 \alpha_j}{\sum_{i=1}^k \alpha_i}-\frac{R^2+G^2\sum_{i=1}^k \alpha_i^2}{2\Big(\sum_{i=1}^k \alpha_i\Big)^2}$$ Since the same second term appears in all derivatives, then the first term is constant which implies that, when all derivatives are equal, all the $\alpha_j$ are the same.

Pushing the calculations to the end, all derivatives will be zero when all the $\alpha$'s will be equal to $\pm\frac {R}{G \sqrt k}$ and the corrsponding value of $F$ should be $\pm\frac{G R}{\sqrt{k}}$.

Answer 3

Looking at the definition of $f$ from a statistical standpoint, define the average value of the $\alpha_i$'s to be $\bar\alpha= \frac{1}{k}\sum_{i=1}^k \alpha_i$.

Define the variance of the $\alpha_i$'s to be $\sigma^2=\frac{1}{k}\sum_{i=1}^k (\alpha_i - \bar \alpha)^2= \frac{1}{k}(\sum_{i=1}^k \alpha_i^2 - k\bar\alpha^2)$.

Then $f$ can be redefined in terms of $\bar \alpha$ and $\sigma^2$ as

$$f(\alpha) = \frac{R^2+G^2(k\sigma^2 + k\bar\alpha^2)}{2k\bar\alpha}$$ where by definition all terms in the numerator are non-negative and the denominator is positive.

For any given value of $\bar \alpha$ (or equivalently, any given value of $\sum_{i=1}^k \alpha_i$) notice that the numerator of $f$, and hence $f$ itself, is minimized when the variance $\sigma^2$ of the $\alpha_i$'s is zero. This happens precisely when all of the $\alpha_i$'s are equal to their mean value and hence are identical.

Under the condition that the variance is zero, $f$ could be differentiated with respect to the mean of the $\alpha_i$'s which is now just a single number $\alpha$ to find its value that further minimizes $f$ in terms of R, G, and k.