For positive numbers $a, b, c$ it holds that

$$
(a+c)\left(b^{2}+a c\right)=4 a .
$$

Determine the maximum value of the expression $b+c$ and find all triples of numbers $(a, b, c)$ for which this expression attains this value.

(Michal Rolínek)
Solution. We will skillfully transform and estimate the given equality using the known inequality $a^{2}+b^{2} \geqq 2 a b$ as follows:

$$
4 a=(a+c)\left(b^{2}+a c\right)=a\left(b^{2}+c^{2}\right)+c\left(a^{2}+b^{2}\right) \geqq a\left(b^{2}+c^{2}\right)+2 a b c=a(b+c)^{2} .
$$

From this, we see that $b+c \leqq 2$, and also that equality occurs if and only if $00$. That is all.