$x_1 \leq 4, x_1+x_2 \leq 13, x_1+x_2+x_3 \leq 29, x_1+…+x_4 \leq 54, x_1+…+x_5 \leq 90 \text{ for } x_1, …, x_5 \in R_0^+. \\ \ \\ \text{Find the maximum of } \sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}+\sqrt{x_4}+\sqrt{x_5}.$
Of course, the answer will be $20(x_1=4, x_2=9, x_3=16, x_4=25, x_5=36.)$
My attempt:
\begin{align}
&\text{let } \sum_k \ x_l = x_l+x_{l+1}+x_{l+2}+…+x_{k+l-1}. \ (x_{5n+m}=x_m) \\
&\bigg(\sum_5 \sqrt{x_1} \bigg)^2 \leq 5\bigg(\sum_5 x_1\bigg) \leq 450. \\
&\therefore \bigg( \sum_5 \sqrt{x_1} \bigg) \leq 15\sqrt{2}. \\
\ \\
&\bigg(\sum_4 \sqrt{x_1} \bigg)^2 \leq 4\bigg(\sum_4 x_1 \bigg) \leq 216. \\
&\therefore \bigg( \sum_4 \sqrt{x_1} \bigg) \leq 6\sqrt{6}. \\
\ \\
&\bigg(\sum_3 \sqrt{x_1} \bigg)^2 \leq 3\bigg(\sum_3 x_1 \bigg) \leq 87. \\
&\therefore \bigg(\sum_3 \sqrt{x_1} \bigg) \leq \sqrt{87}. \\
\ \\
&\bigg(\sum_2 \sqrt{x_1} \bigg)^2 \leq 2\bigg(\sum_2 x_1 \bigg) \leq 26. \\
&\therefore \bigg(\sum_2 \sqrt{x_1} \bigg) \leq \sqrt{26}. \\
\ \\
&\bigg(\sqrt{x_1} \bigg)^2 \leq 4. \\
&\therefore \sqrt{x_1} \leq 2. \\
\end{align}
Best Answer
$$x_1 \leq 4 \tag {1}$$ $$x_1+x_2 \leq 13\tag{2}$$ $$x_1+x_2+x_3 \leq 29\tag{3}$$ $$x_1+...+x_4 \leq 54\tag {4}$$ $$x_1+...+x_5 \leq 90\tag 5$$
now $$(5)\times 10+(4)\times 2+(3)\times 3+(2)\times 5+(1)\times 10 \implies$$ $$10x_5+12x_4+15x_3+20x_2+30x_1\le 1200$$ also by C-S $${\left(10x_5+12x_4+15x_3+20x_2+30x_1 \right)}{\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{15}+\frac{1}{20}+\frac{1}{30}\right)}\ge {\left(\sum \sqrt{x_i}\right)}^2$$ can you end it now?