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} One may note that $g(1)=1$. 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 $p_{n,j}:=c_j c_{n+1-j}$. Indeed, then for $h:=fg$ one has $h(z)=\frac1{\sqrt2}\sum_{n=0}^\infty s_n z^n$, where $s_0:=a_0b_0=1/2>0$.

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.

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} One may note that $g(1)=1$. 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 $p_{n,j}:=c_j c_{n+1-j}$. Indeed, then for $h:=fg$ one has $h(z)=\frac1{\sqrt2}\sum_{n=0}^\infty (b_n+s_n) z^n$, where $s_0:=a_0b_0=1/2>0$.

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.

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.

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} One may note that $g(1)=1$. 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 $p_{n,j}:=c_j c_{n+1-j}$. Indeed, then for $h:=fg$ one has $h(z)=\frac1{\sqrt2}\sum_{n=0}^\infty s_n z^n$, where $s_0:=a_0b_0=1/2>0$. 

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.