3. It is known that for positive numbers $a, b$, and $c$, each of the three equations $a x^{2} +$ param $1 b x + c = 0$, $b x^{2} +$ param $1 c x + a = 0$, $c x^{2} +$ param $1 a x + b = 0$ has at least one real root.

What is the smallest possible value for the product of the roots of the second equation if the product of the roots of the first equation is equal to param2? (If the equation has two coinciding roots, the product is considered to be the square of this root).

| param1 | param2 | Answer |
| :---: | :---: | :---: |
| 10 | 6 | 0.24 |
| 20 | 7 | 0.07 |
| 12 | 9 | 0.25 |
| 8 | 14 | 0.875 |
| 15 | 18 | 0.32 |

## Condition

It is known that for positive numbers $a, b, c$, each of the three equations

$a x^{2} + 10 b x + c = 0$,

$b x^{2} + 10 c x + a = 0$,

$c x^{2} + 10 a x + b = 0$

has at least one real root.

What is the smallest possible value for the product of the roots of the second equation if the product of the roots of the first equation is equal to 6? (If the equation has two coinciding roots, the product is considered to be the square of this root).

## Answer.

0.24
# Solution

Let $10 / 2=p, 6=q$. By Vieta's formulas, $c / a=q$. The discriminant of the third equation is non-negative, so $a^{2} p^{2} \geq b c$. Replacing $c$ in this inequality, we get: $a^{2} p^{2} \geq a b q$. Therefore, $a / b \geq q / p^{2}$. Hence, $\min \{a / b\}=q / p^{2}=0.24$, which is the minimum value of the product of the roots of the second quadratic. It remains to check that this minimum is achieved, for example, by taking $a=q, b=p^{2}, c=q^{2}$.