If $a+b+c=1$ and $abc>0$, then $ab+bc+ac<\frac{\sqrt{abc}}{2}+\frac{1}{4}.$ Question:

For any $a,b,c\in \mathbb{R}$ such that $a+b+c=1$ and $abc>0$, show that
  $$ab+bc+ac<\dfrac{\sqrt{abc}}{2}+\dfrac{1}{4}.$$

My idea: let 
$$a+b+c=p=1, \quad ab+bc+ac=q,\quad abc=r$$
so that
$$\Longleftrightarrow q<\dfrac{\sqrt{r}}{2}+\dfrac{1}{4}$$
Note this $a,b,c\in \mathbb{R}$, so we can't use schur inequality such
$$p^3-4pq+9r\ge 0, \quad pq\ge 9r$$
and so on
maybe can use AM-GM inequality to solve it.

Edit: Incomplete approach. Only works if $a,b,c\geq 0$.
By replacing $\frac{1}{4}$ on the RHS with $\frac{(a+b+c)^2}{4}$, the inequality you seek is equivalent to
$$
a^2+b^2+c^2+2\sqrt{abc}>2(ab+bc+ca).\tag{I}
$$
To prove (I), we use the following result
$$
a^2+b^2+c^2+3(abc)^{2/3}\geq 2(ab+bc+ca)\tag{II}
$$
the proof of which can be found here. Because of (II), it is enough to verify now that
$$
2\sqrt{abc}>3(abc)^{2/3}\iff abc<\left(\frac{2}{3}\right)^6
$$
but this last inequality follows from the AM-GM inequality
$$
\sqrt[3]{abc}\leq\frac{a+b+c}{3}=\frac{1}{3}\implies abc\leq\left(\frac{1}{3}\right)^3<\left(\frac{2}{3}\right)^6.
$$
