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

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

Solution 2. The given equation is equivalent to

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

Now observe that by the well known inequality \(a^{2}+b^{2}+c^{2} \geq a b+b c+c a\), with \(\frac{1}{y+z}, \frac{1}{z+x}\), \(\frac{1}{x+y}\) in place of \(a, b, c\); we get:

\[
\begin{aligned}
\frac{27}{4} & =(x+y+z)^{2}\left(\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}+\frac{1}{(x+y)^{2}}\right) \\
& \geq(x+y+z)^{2}\left(\frac{1}{(y+z)(z+x)}+\frac{1}{(z+x)(x+y)}+\frac{1}{(x+y)(y+z)}\right)=\frac{2(x+y+z)^{3}}{(x+y)(y+z)(z+x)} \\
& =\frac{(2(x+y+z))^{3}}{4(x+y)(y+z)(z+x)}=\frac{((x+y)+(y+z)+(z+x)))^{3}}{4(x+y)(y+z)(z+x)} \stackrel{\text { AM-GM }}{\geq \frac{(3 \sqrt[3]{(x+y)(y+z)(z+x)})^{3}}{4(x+y)(y+z)(z+x)} .} \\
& =\frac{27}{4}
\end{aligned}
\]

This means all inequalities in the above calculations are equalities, and this holds exactly whenever \(x+y=y+z=z+x\), that is \(x=y=z\). By the statement's demand we need to have \(a, b, c\) integers. And conversely, any triad of equal natural numbers \(x=y=z\) is indeed a solution for the given equation, and so these are all its solutions.