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.

OK. Let me try to complete the answer.
We want to prove that for $a,b,c\in \mathbb{R},abc>0$,
$$ab+bc+ac<\dfrac{\sqrt{abc}}{2}+\dfrac{1}{4}. \tag{1}$$
In a previous post, it is proved for $a,b,c\ge 0$
Because $abc>0$, the only other possibility is that two of the numbers are negative and one is positive. We can assume that $a=A>0,-b=B>0,-c=C>0$.
It is then suffice to prove that 
$$ab+bc+ac=AB-C(A+B)=AB-(1+A+B)(A+B)<0 \tag{2}$$
Because 
$$(1+A+B)(A+B)>(A+B)(A+B)\ge 4AB > AB \tag{3}$$
We know that (2) is true.
