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. 

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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 $a_n=0$.