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.
 A: 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.
A: 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
$$
\sqrt[3]{abc}\leq\frac{a+b+c}{3}=\frac{1}{3}\implies abc\leq\left(\frac{1}{3}\right)^3<\left(\frac{2}{3}\right)^6.
$$
A: At least one of $a, b, c$ must be positive. Assume $a > 0$. Then we have $a^2 + ab + ac = a(a + b + c) = a$, so the inequality to be proved is equivalent to
$$bc + a - a^2 < \frac{1}{2}\sqrt{abc} + \frac{1}{4},$$
which can be rewritten as $f(\sqrt{bc}) < 0$, where 
$$f(x) = x^2 - \frac{1}{2}\sqrt{a}x - a^2 + a - \frac{1}{4}.$$ 
Now $b$ and $c$ are subject only to the conditions $b + c = 1 - a$ and $bc > 0$. Thus $\sqrt{bc}$ varies in the interval $(0,\frac{1}{2}|1-a|]$.
Since $f(x)$ is a quadratic polynomial with positive leading coefficient, to check the inequality, it's enough to check it at its endpoints, namely, $f(0) \leq 0$ and $f(\frac{1}{2}|1-a|) < 0$. But $f(0) = -(a-\frac{1}{2})^2$, so the first of these inequalities is clear. Therefore, we need only prove that for all $a > 0$, we have
$$\frac{1}{4}(a-1)^2 -\frac{1}{4}\sqrt{a}|1-a| -a^2 +a - \frac{1}{4} < 0,$$
which we can rewrite as 
$$-\frac{3}{4}a^2 +\frac{1}{2}a - \frac{1}{4}\sqrt{a}|1-a| < 0.$$
As the sum of the first two terms is already negative for $a \geq 1$, we need only consider the case $0 < a < 1$. Dividing by $t = \frac{1}{4}\sqrt{a}$, the inequality becomes
$$-3t^3 + t^2 + 2t - 1 < 0,$$
which must be proved for all $t \in (0,1)$ in order to conclude the proof.
This is straightforward to do with calculus.
