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.

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.
