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 $x^{2}y+x{y}^2+y^{2}z+y{z}^2+x^{2}z+x{z}^2\ge 6xyz$. 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).