Manually find the maximum values of $\sqrt{x_i}$ and add them together?

We' ve got $\sqrt{x_1} \leq \dfrac{x_1 +4}{4} (1)$ $\sqrt{x_2} \leq \dfrac{x_2+9}{6} (2)$ $\sqrt{x_3} \leq \dfrac{x_3+16}{8} (3)$ $\sqrt{x_4} \leq \dfrac{x_4+25}{10} (4)$ $\sqrt{x_5} \leq \dfrac{x_5+36}{12} (5)$ $(1)+(2)+(3)+(4)+(5) \Rightarrow \sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}+\sqrt{x_4}+\sqrt{x_5}\leq 10+ \dfrac{x_1+x_2+x_3+x_4+x_5}{12}+\dfrac{x_1+x_2+x_3+x_4}{60}+\dfrac{x_1+x_2+x_3}{40}+\dfrac{x_1+x_2}{24}+\dfrac{x_1}{12}\leq 20$

Claim. Suppose $\sum_{j=1}^i \sqrt{x_j}\leq \frac{i^2+3i}{2}$ for all $1\leq i\leq n-1$ and $\sum_{j=1}^i x_j\leq \frac{(i+1)(i+2)(2i+3)-6}{6}$ for all $1\leq i\leq n$. Then $$\sum_{j=1}^n \sqrt{x_j}\leq \frac{n^2+3n}{2}.$$Proof. Note that by Power Mean, \begin{align*} n(n+1)^2&= \frac{(n-1)n(2n-1)}{6}+\frac{(n+1)(n+2)(2n+3)-6}{6}+ \frac{3(n-1)n}{2}+\frac{(n-1)n(2n-1)}{6}\\&\geq \frac{(n-1)n(2n-1)}{6}+\frac{(n+1)(n+2)(2n+3)-6}{6}+ \sum_{j=1}^{n-1} (j^2+3j)\\&\geq \sum_{j=1}^n (n-j)^2+\sum_{j=1}^n x_j+ \sum_{j=1}^n 2(n-j)\sqrt{x_j}\\&= \sum_{j=1}^n (\sqrt{x_j}+n-j)^2\\&\geq \frac{(\sqrt{x_n}+\sqrt{x_{n-1}}+1+\ldots+\sqrt{x_1}+n-1)^2}{n}\\&=\frac{\sum_{j=1}^n (\sqrt{x_j}+n-j)^2}{n} \\&=\frac{\sum_{j=1}^n \sqrt{x_j}+\frac{(n-1)n}{2})^2}{n}. \end{align*}Therefore, $$\frac{n^2+3n}{2}=n(n+1)-\frac{(n-1)n}{2}\geq \sum_{j=1}^n \sqrt{x_j}.$$$\square$ We are done by the claim applied together with an induction argument.

$$x_1+x_2+x_3+x_4+x_5\leq 90\Rightarrow 10(x_1+x_2+x_3+x_4+x_5)\leq 900$$$$x_1+x_2+x_3+x_4\leq 54\Rightarrow (12-10)(x_1+x_2+x_3+x_4)\leq 54(12-10)=108$$$$x_1+x_2+x_3\leq 29\Rightarrow (15-12)(x_1+x_2+x_3)\leq 29(15-12)=87$$$$x_1+x_2\leq 13\Rightarrow (20-15)(x_1+x_2)\leq 13(20-15)=65$$$$x_1\leq 4\Rightarrow (30-20)(x_1)\leq 4(30-20)=40$$Sum up these inequalities side by side: $$30x_1+20x_2+15x_3+12x_4+10x_5\leq 1200$$$$\left(\frac 1{30}+\frac 1{20}+\frac 1{15}+\frac 1{12}+\frac 1{10}\right)(30x_1+20x_2+15x_3+12x_4+10x_5)\leq 1200\left(\frac 1{30}+\frac 1{20}+\frac 1{15}+\frac 1{12}+\frac 1{10}\right)=400$$$$400\ge \left(\frac 1{30}+\frac 1{20}+\frac 1{15}+\frac 1{12}+\frac 1{10}\right)(30x_1+20x_2+15x_3+12x_4+10x_5)\ge (\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}+\sqrt{x_4}+\sqrt{x_5})^2$$$$20\ge \sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}+\sqrt{x_4}+\sqrt{x_5}$$