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