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

Question

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.

Answer 1

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.

Answer 2

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 $$ \sqrt[3]{abc}\leq\frac{a+b+c}{3}=\frac{1}{3}\implies abc\leq\left(\frac{1}{3}\right)^3<\left(\frac{2}{3}\right)^6. $$

Answer 3

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.