The equation with integer coefficients $x^{4}+a x^{3}+b x^{2}+c x+d=0$ has four positive roots, counting multiplicities.

Find the smallest possible value of the coefficient $b$ under these conditions.
Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the roots of the equation (some of them may coincide). By Vieta's theorem, $b=x_{1} x_{2}+x_{1} x_{3}$ $+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}$

$d=x_{1} x_{2} x_{3} x_{4}$, hence $b$ and $d$ are positive. Notice that

$$
\frac{b}{\sqrt{d}}=\left(\sqrt{\frac{x_{1} x_{2}}{x_{3} x_{4}}}+\sqrt{\frac{x_{3} x_{4}}{x_{1} x_{2}}}\right)+\left(\sqrt{\frac{x_{1} x_{3}}{x_{2} x_{4}}}+\sqrt{\frac{x_{2} x_{4}}{x_{1} x_{3}}}\right)+\left(\sqrt{\frac{x_{1} x_{4}}{x_{2} x_{3}}}+\sqrt{\frac{x_{2} x_{3}}{x_{1} x_{4}}}\right) \geq 2+2+2 \text { (inequality }
$$

of Koshi). Therefore, $b \geq 6$ ( $d$ is an integer, so $d \geq 1$ ). Equality is achieved when the equation has four repeated roots, all equal to 1. In this case, the polynomial has the form

$(x-1)^{4}=x^{4}-4 x^{3}+6 x^{2}-4 x+1$

## Answer

$b=6$.