Hello, it is $$b+c\le 2$$and the equal sign holds if $$a=b=c=1$$Sonnhard.

parmenides51 wrote: For positive numbers $a, b, c$ holds $(a + c) (b^2 + a c) = 4a$. Determine the maximum value of $b + c$ and find all triplets of numbers $(a, b, c)$ for which expression takes this value From Cauchy-Schwarz: $4=\left(1+\frac{c}{a}\right)(b^2+ca)\geq (b+c)^2$ so $b+c\leq 2$. Equality holds when $(a,b,c)=(t,t,2-t)$ with $t\in (0,2)$.