KöMaL 2026 A933

Let $a$, $b$, $c$ be real numbers in the interval $(1,4)$. Consider the following three inequalities: ${bx(x+y-z)}+{cx(x+z-y)}\geq {a(2yz+x)}$, ${cy(y+z-x)}+{ay(y+x-z)} \geq {b(2zx+y)}$, ${az(z+x-y)}+{bz(z+y-x)} \geq {c(2xy+z)}$. a) Determine, in terms of $a$, $b$, $c$, the maximum number of these inequalities that can hold simultaneously for positive real numbers $x$, $y$, $z$ satisfying $x+y+z=1$. b) Find an explicit formula for $x$, $y$, $z$ in terms of $a$, $b$, $c$, without using case distinctions, such that $x$, $y$, $z>0$, $x+y+z=1$, and none of the three inequalities holds.