TITLE: If $a+b+c=1$ and $abc>0$, then $ab+bc+ac<\frac{\sqrt{abc}}{2}+\frac{1}{4}.$
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: WLOG, let $a$ be the largest, so $a> 0 \implies 1> b+c \implies a> ab+ca$.  So it is sufficient to show that
$$4(a+bc) \le 2\sqrt{abc}+1$$
Look at this as a quadratic in $\sqrt{bc}$ and then it is enough to show the discriminant is never positive.
$$\Delta = 4a - 16(4a-1) \le 0 \iff 3a \ge 1$$
which is true as $a$ is the largest.