
A4. Solve the following equation for $x, y, z \in \mathbb{N}$

$$
\left(1+\frac{x}{y+z}\right)^{2}+\left(1+\frac{y}{z+x}\right)^{2}+\left(1+\frac{z}{x+y}\right)^{2}=\frac{27}{4}
$$


Solution 1. Call $a=1+\frac{x}{y+z}, b=1+\frac{y}{z+x}, c=1+\frac{z}{x+y}$ to get

$$
a^{2}+b^{2}+c^{2}=\frac{27}{4}
$$

Since it is also true that

$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2
$$

the quadratic-harmonic means inequality implies

$$
\frac{3}{2}=\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{3}{2}
$$

So the inequality in the middle holds as an equality, and this happens whenever $a=b=c$, from which $1+\frac{x}{y+z}=1+\frac{y}{z+x}=1+\frac{z}{x+y}$.

But $1+\frac{x}{y+z}=1+\frac{y}{z+x} \Leftrightarrow x^{2}+x z=y^{2}+y z \Leftrightarrow(x-y)(x+y)=z(y-\dot{x})$ and the two sides of this equality will be of different sign, unless $x=y$ in which case both sides become 0 . So $x=y$, and similarly $y=z$, thus $x=y=z$.

Indeed, any triad of equal natural numbers $x=y=z$ is a solution for the given equation, and so these are all its solutions.
