Positive reals satisfy $ \sum_{i=1}^{24} x_i = 1 $, determine maximum of following quantity

Question

So, positive reals satisfy the following

$$ \sum_{i=1}^{24} x_i = 1 $$

And I need to find maximum of the following quantity.

$$ \left( \sum_{i=1}^{24} \sqrt{x_i}\right) \left(\sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i} } \right) $$

Now, using Cauchy Schwarz inequality, I got

$$ \left( \sum_{i=1}^{24} \sqrt{x_i}\right)^2 \leqslant \underbrace{(1+1+\cdots + 1)}_{\text{24 times}} \left( \sum_{i=1}^{24} x_i \right) $$

This leads to

$$ \left( \sum_{i=1}^{24} \sqrt{x_i}\right) \leqslant \sqrt{24} $$

I am stuck with other part. I can get the minimum of the following using similar technique.

$$ \left(\sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i} } \right) $$

But I need to have maximum of this quantity, so that I can combine the two. Any hints will help.

Answer 1

We can bound the second sum as follows. Using Cauchy-Schwarz inequality, we have the following.

$$ \left( \sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i}} \right)^2 \leqslant \left( \sum_{i=1}^{24} \frac{1}{(1+x_i)} \right)\underbrace{(1+1+\cdots +1)}_{\text{24 times}} $$

$$ \left( \sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i}} \right)^2 \leqslant 24 \left( \sum_{i=1}^{24} \frac{1}{(1+x_i)} \right) \, \cdots \cdots \cdots(1) $$

Now, I will use Hölder's inequality.

$$ \left( \sum_{i=1}^{24} \frac{1}{(1+x_i)} \right)^{1/2} \left( \sum_{i=1}^{24} (1+x_i) \right)^{1/2} \leqslant \left[ \sum_{i=1}^{24} \left(\frac{1}{\sqrt{1+x_i}}\right) \left(\sqrt{1+x_i}\right) \right] $$

$$ \left( \sum_{i=1}^{24} \frac{1}{(1+x_i)} \right)^{1/2} \sqrt{25} \leqslant 24 $$

$$ \left( \sum_{i=1}^{24} \frac{1}{(1+x_i)} \right) \leqslant \frac{24^2}{25} $$

$$ 24 \left( \sum_{i=1}^{24} \frac{1}{(1+x_i)} \right) \leqslant \frac{24^3}{25} $$

So, combining with equation $(1)$, I get,

$$ \left( \sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i}} \right)^2 \leqslant \frac{24^3}{25} $$

$$ \left( \sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i}} \right) \leqslant \frac{24^{3/2}}{5} $$

Finally, combining the two sums, I get

$$ \left( \sum_{i=1}^{24} \sqrt{x_i} \right) \left( \sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i}} \right) \leqslant \sqrt{24} \,\frac{24^{3/2}}{5} $$

$$ \left( \sum_{i=1}^{24} \sqrt{x_i} \right) \left( \sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i}} \right) \leqslant \frac{24^{2}}{5} $$

Hope that helps