Rewrite your inequality as $$lhs:=\sqrt{s_1}+\sum_{n=2}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}\le C\sqrt{s_N},$$ where $s_n:=\sum_{i=1}^n a_i$. Note that $$\sum_{n=2}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}$$ is a lower Riemann sum for the integral $$\int_{s_1}^{s_N}\frac{ds}{\sqrt s}\le2\sqrt{s_N}.$$ So, $$lhs\le\sqrt{s_1}+2\sqrt{s_N}\le3\sqrt{s_N},$$ as desired.

In the above proof, it was tacitly assumed that $a_i>0$ for all $i$. This can be obviously extended to the case when we only know that $a_i\ge0$ for all $i$ -- assuming that, by continuity, $\frac{a_n}{\sqrt{\sum_{i=1}^{n}a_i}}:=0$ whenever $a_n=0$.

Rewrite your inequality as $$lhs:=\sum_{n=1}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}\le C\sqrt{s_N},$$ where $s_n:=\sum_{i=1}^n a_i$. Note that $$\sum_{n=1}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}$$ is a lower Riemann sum for the integral $$\int_0^{s_N}\frac{ds}{\sqrt s}=2\sqrt{s_N}.$$ So, $$lhs\le2\sqrt{s_N},$$ as desired.

In the above proof, it was tacitly assumed that $a_i>0$ for all $i$. This can be obviously extended to the case when we only know that $a_i\ge0$ for all $i$ -- assuming that, by continuity, $\frac{a_n}{\sqrt{\sum_{i=1}^{n}a_i}}:=0$ whenever $a_n=0$.

Rewrite your inequality as $$lhs:=\sqrt{s_1}+\sum_{n=2}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}\le Cs_N,$$ where $s_n:=\sum_{i=1}^n a_i$. Note that $$\sum_{n=2}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}$$ is a lower Riemann sum for the integral $$\int_{s_1}^{s_N}\frac{ds}{\sqrt s}\le2\sqrt{s_N}.$$ So, $$lhs\le\sqrt{s_1}+2\sqrt{s_N}\le3\sqrt{s_N},$$ as desired.

Rewrite your inequality as $$lhs:=\sqrt{s_1}+\sum_{n=2}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}\le C\sqrt{s_N},$$ where $s_n:=\sum_{i=1}^n a_i$. Note that $$\sum_{n=2}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}$$ is a lower Riemann sum for the integral $$\int_{s_1}^{s_N}\frac{ds}{\sqrt s}\le2\sqrt{s_N}.$$ So, $$lhs\le\sqrt{s_1}+2\sqrt{s_N}\le3\sqrt{s_N},$$ as desired.

In the above proof, it was tacitly assumed that $a_i>0$ for all $i$. This can be obviously extended to the case when we only know that $a_i\ge0$ for all $i$ -- assuming that, by continuity, $\frac{a_n}{\sqrt{\sum_{i=1}^{n}a_i}}:=0$ whenever $a_n=0$.