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Let $0 \leq x_1, \dots, x_n \leq 1$ and $\sum_{i=1}^n x_i = 1$, i.e. $x_i$ form a probability distribution. Let $m_1, \dots, m_n \geq 1$ be fixed numbers. I'm looking for $x_i$ that maximize the following:

\begin{align} \sum_i \frac{\sqrt{x_i}}{\sqrt{m_i}} \end{align}

I was able to show that for case of $n = 2$ (where we only have a single variable) the maximum is at $x_1 = \frac{m_2}{m_1+m_2}$ and $x_1 = 1-x_2$. Let $M = \sum_i m_i$. From this, I think $x_i$ that maximize the above might be

$$x_i \propto {M - m_i}?$$

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    $\begingroup$ Your guess isn't true in general, because then $\sum x_i$ would equal $n-1$, which is only equal to $1$ if $n = 2$. $\endgroup$ May 19, 2021 at 0:21
  • $\begingroup$ @VarunVejalla Right! I fixed my question to reflect that. $\endgroup$
    – KRL
    May 19, 2021 at 0:26
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    $\begingroup$ I would have said your $n=2$ example suggests $x_1 m_1 = x_2m_2$ leading to the guess $x_i \propto \frac1{m_i}$ $\endgroup$
    – Henry
    May 19, 2021 at 0:38

2 Answers 2

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I think the most straightforward way to solve this problem would be with Lagrange multipliers. The Lagrangian would be

$$L(x, \lambda) = \sum_i \frac{\sqrt{x_i}}{\sqrt{m_i}} - \lambda\left(\sum_i x_i - 1\right)$$

Then taking the derivatives with respect to the $x_i$, you'd need $$\frac{1}{2\sqrt{x_i}\sqrt{m_i}}-\lambda=0$$ for all $i$. This can be solved for $x_i$ as $$x_i=\frac{1}{4\lambda^2 m_i}$$ and since $\sum_i x_i$ must equal $1$, you have that $$\sum_i \frac{1}{4\lambda^2 m_i} = 1 \to \lambda = \frac{1}{2}\sqrt{\sum_i \frac{1}{m_i} }$$

So $$x_i = \frac{1}{m_i\sum_{i}\frac{1}{m_i} }$$ which yields a maximum of $$\sqrt{\sum_i \frac{1}{m_i}}$$

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Apply the Cauchy-Schwarz inequality $$\left(\sum \frac{\sqrt{x_i}}{\sqrt{m_i}}\right)^2 \le \sum \frac{1}{m_i} \cdot \sum x_i $$ with equality only when $x_i$ and $\frac{1}{m_i}$ are proportional.

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