Bulgarian Spring Tournament

a) The inequality $(xy+yz+zx)(x+y+z) \ge 1 = (x+y)(y+z)(z+x)$ is equivalent to $xyz \ge 0$. Equality is reached only when one of the variables, say $x$, is $0$ and the other two (in this case $y$ and $z$) are whatever with $yz(y+z) = 1$; one possibility is $y=1$ and $z = \frac{\sqrt{5}-1}{2}$. The inequality $(xy+yz+zx)(x+y+z) \le \frac{9}{8}$ is equivalent to $9(x+y)(y+z)(z+x) - 8(xy+yz+zx)(x+y+z) \ge 0$, i.e. on $x^2y + xy^2 + y^2z + yz^2 + x^2z + xz^2 \ge 6xyz$. The latter is true from the inequality between the arithmetic mean and the geometric mean applied to the six terms on the left, with equality only at $x = y = z$ and $8x^3 = 1$, i.e. $x = y = z = \frac{1}{2}$.

b) Given the reasoning in a), it suffices to prove that the equation $yz(y+z) = 1$ has no solution in (positive) rational numbers. Assume the opposite and let $y = \frac{p}{r}$, $z = \frac{q}{r}$ is a solution (with natural $p, q, r$) in which we have reduced the fractions under a common denominator. Then $pq(p+q) = r^3$, as after truncating a common divisor of $p, q, r$, if necessary, we can consider that $(p, q, r) = 1$. In fact, here already $(p, q) = 1$, since otherwise their common prime divisor would also be that of $r$, contradiction with $(p, q, r) = 1$. Thus, the numbers $p, q$ and $p+q$ are two by two mutually prime, and since their product is an exact cube, then necessarily $p = a^3$, $q = b^3$ and $p+q = c^3$ for some natural numbers $a, b, c$. Thus we obtained $a^3 + b^3 = c^3$ for some natural $a, b, c$, which is impossible (a special case of the so-called Fermat's Great Theorem).