Posed in regional mathematics Olympiad 2005 Let a, b, c be three positive real numbers such that $a+b+c = 1$.
Let $\Delta= \min( a^{3} + a^{2}bc, b^{3}+ab^{2}c, c^{3}+abc^{2} )$.
Prove that the roots of the equation $x^{2} + x + 4 \Delta = 0$ are real.
The last line is equivalent to $\Delta \leq \frac{1}{16} $. So I tried to prove by contradiction. I assumed all of them are $ > \frac{1}{16} $and tried to draw a contradiction  with the fact $a+ b+ c=1$ but failed

First, we see that the roots of the polynimium are real iff:
$$1-16\Delta \ge 0 \Leftrightarrow \Delta \le \frac{1}{16}$$
Assume without loss of gennerality that $a \le b \le c$. Then:
$$\Delta = a^3 + a^2bc = a^2(a+bc)$$
$bc$ is biggest when $b=c=\frac{1-a}{2}$, so:
$$\Delta\le a^2\left(a+\left(\frac{1-a}{2}\right)^2\right)=\frac{a^4+2a^3+a^2}{4}$$
Notice that this expression grows monotonically with $a$, but $a\le \frac13$, so it's maximum is at $a=\frac13$. But then:
$$\Delta\le\frac{\frac{1}{81}+\frac{2}{27}+\frac{1}{9}}{4}=\frac{4}{81} < \frac{1}{16}$$ 
