Consider the polynomials $p(x)=x^{4}+a x^{3}+b x^{2}+c x+1$ whose coefficients are positive numbers less than 3, and which have no real roots. What is the maximum possible value of $a b c$?
Since $p$ is a continuous function and $p(0)=1>0$, if $p(-1) \leq 0$, then by the Bolzano theorem, $p$ takes the value 0 on the interval $[-1,0]$. But $p$ has no real roots, so $p(-1)=1-a+b-c+1>0$, from which $a+c<b+2<5$. Therefore, $(a+c)^{2}<25$. Expanding the parentheses and using the inequality $2 a c \leq a^{2}+c^{2}$, we get $a c<6.25$. Given the condition $b<3$, we have $a b c<18.75$. Thus, if $a, b, c$ are positive numbers less than 3 and $p$ has no roots, then $a b c<18.75$.

We will also show that $a b c$ can take any value in the interval $[0 ; 18.75)$. Let $a=c=2.5-\varepsilon$ and $b=3-\frac{\varepsilon}{2}$. If $\varepsilon$ moves in the interval $(0 ; 2.5)$, then $a b c$ takes every value in the interval $(0 ; 18.75)$ by the Bolzano theorem.

Now we only need to prove that in this case, $p$ has no real roots. Assume the contrary:

$$
x^{4}+(2.5-\varepsilon) x^{3}+\left(3-\frac{\varepsilon}{2}\right) x^{2}+(2.5-\varepsilon)+1=0
$$

Since $x \neq 0$, we can divide by $x^{2}$:

$$
x^{2}+\frac{1}{x^{2}}+(2.5-\varepsilon)\left(x+\frac{1}{x}\right)+\left(3-\frac{\varepsilon}{2}\right)=0
$$

Rearranging:

$$
\left(x+\frac{1}{x}\right)^{2}+(2.5-\varepsilon)\left(x+\frac{1}{x}\right)+\left(1-\frac{\varepsilon}{2}\right)=0
$$

Using the quadratic formula for $x+\frac{1}{x}$, we get:

$$
\frac{\varepsilon-2.5 \pm \sqrt{\varepsilon^{2}-3 \varepsilon+2.25}}{2}
$$

Under the square root, we have $(\varepsilon-1.5)^{2}$, so $x+\frac{1}{x}=-\frac{1}{2}$ or $x+\frac{1}{x}=\varepsilon-2$.

Multiplying both equations by $x$, we get equations that have no real roots. Therefore, we have reached a contradiction.

Zoltán Szabó (Veszprém, L. Lovassy Gymnasium, 12th grade)