(a) Determine the largest possible value $M$ that $x+y+z$ can take if $x$, $y$, and $z$ are positive real numbers such that

$$
16 x y z=(x+y)^{2}(x+z)^{2}
$$

(b) Show that there are infinitely many triples $(x, y, z)$ of positive rational numbers for which

$$
16 x y z=(x+y)^{2}(x+z)^{2} \text { and } x+y+z=M
$$

hold.

(Karl Czakler)
(a) Due to the side condition and the arithmetic-geometric mean inequality, we have

$$
4 \sqrt{x y z}=(x+y)(x+z)=x(x+y+z)+y z \geq 2 \sqrt{x y z(x+y+z)} .
$$

Thus, $2 \geq \sqrt{x+y+z}$ and therefore $4 \geq x+y+z$. Since in the second part we provide infinitely many such triples for which $x+y+z=4$, $M=4$ is the desired maximum.
(b) In the equality case, equality must hold in the mean inequality in the first part, i.e., $x(x+y+z)=y z$, and of course also $x+y+z=4$. If we choose $y=t$, with $t$ rational, we get $4 x=t(4-x-t)$ and thus $x=\frac{4 t-t^{2}}{4+t}$ and $z=4-x-y=\frac{16-4 t}{4+t}$. If $0<t<4$ also holds, then these expressions are all positive and rational, and the side condition is also satisfied.

The triples $\left(\frac{4 t-t^{2}}{4+t}, t, \frac{16-4 t}{4+t}\right)$ with $0<t<4$ rational are thus infinitely many equality cases.

(Karl Czakler)