Let $a_j:=\binom{1/2}j$. Then $a_0=1$ and $a_j=(-1)^{j-1}c_j$ for $j=1,2,\dots$, where 
\begin{equation}
	c_j:=\frac1{j2^{2j-1}} \binom{2 j-2}{j-1}>0. 
\end{equation}
Let 
\begin{equation}
	g(z):=\sum_{j=0}^\infty b_j z^j,\quad \text{where}\quad 
b_j:=c_{j+1}>0. 
\end{equation}
Since $a_0=1>0$ and $b_j>0$ for all $j\ge0$, it is enough to show that for all natural $n$ 
\begin{equation}
	s_n:=\sum_{j=1}^n a_j b_{n-j}=\sum_{j=1}^n (-1)^{j-1}p_{n,j}\overset{\text{?}}\ge0,
\end{equation}
where $p_{n,j}:=c_j c_{n+1-j}$. Obviously, $p_{n,j}=p_{n,n+1-j}$. So, $s_n=0$ if $n$ is even. 

Let now $n=2m+1$ be odd, so that $m\in\{0,1,\dots\}$. Then 
\begin{equation}
	s_n=s_{2m+1}\ge\sum_{0\le i\le(m-1)/2}(p_{2m+1,2i+1}-p_{2m+1,2i+2}). 
\end{equation}
So, it suffices to show that $p_{2m+1,j}\ge p_{2m+1,j+1}$ for $j=1,\dots,m$. But 
\begin{equation}
	\frac{p_{2m+1,j+1}}{p_{2m+1,j}}-1=-\frac{3 + 6 (m - j)}{(1 + j) (4 m + 1 - 2 j)}<0
\end{equation}
for $j=1,\dots,m$. This completes the proof.