By C-S we have $\frac{81}{4}=3[\left (1+\frac{x}{y+z}\right )^2+\left (1+\frac{y}{z+x}\right )^2+\left (1+\frac{z}{x+y}\right )^2]\ge (3+\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y})^2$.By Nesbitt we have that $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \frac{3}{2}$,so $(3+\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y})^2\ge (3+\frac{3}{2})^2=\frac{81}{4}$. Therefore we must have $x=y=z$ and $x,y,z>0$.

Why $x , y , z \in \mathbb{N}$ and not $\mathbb{R}_+^*$ ? $1+ \frac{y}{z+x} = \frac{x+y+z}{z+x}$ et. al., so $\textrm{LHS} = (x+y+z)^2 \sum \dfrac {1} {(x+y)^2}$. By Cauchy-Schwarz $ \sum \dfrac {1} {(x+y)^2} \geq \dfrac {1}{3} \left ( \sum \dfrac {1} {x+y} \right )^2$. Since $(x+y+z)^2 = \dfrac {1}{4} \left ( \sum (x+y) \right )^2$, we can continue with another Cauchy-Schwarz (avoiding the more occult Nesbitt) \[\textrm{LHS} \geq \dfrac {1}{3\cdot 4} \left ( \left ( \sum (x+y) \right ) \left ( \sum \dfrac {1} {x+y} \right )\right )^2 \geq \dfrac {3^4}{3\cdot 4} = \dfrac {27}{4}.\] Since we are given the equality, that forces $x=y=z$.

mavropnevma wrote: Why $x , y , z \in \mathbb{N}$ and not $\mathbb{R}_+^*$ ? I think people like to do that to confuse people.

The topic title is spoiling , Use successifuly cauchy and nessbit to conclude that it's an equality case of an inequality .

Hi Here's my solution. Put $ \frac {x}{y+z} =a $ and define $ b, c $ similarily. According to Nesbitt we have that $ \sum a \ge \frac {3}{2} $ so using this and simplyfying we have that $ a=b=c= \frac {1}{2} $ so we have $ x=y=z $

Here accually the key was Nesbitt 's inequality!!! Without it you cannot solve the problem, but if someone knows it, he can solve the exercise very quiqly!!! Really nice problem!

Alternative solution: Because the fractions there are homogenous, we can suppose that $x+y+z=S=constant.$ So we hace ${{S}^{2}}\left( {{\left( \frac{1}{S-x} \right)}^{2}}+{{\left( \frac{1}{S-y} \right)}^{2}}+{{\left( \frac{1}{S-z} \right)}^{2}} \right)=\frac{27}{4}$. But the function $f(t)={{\left( \frac{1}{S-t} \right)}^{2}}$, it is convex, so we have Jensen: ${{S}^{2}}\left( {{\left( \frac{1}{S-x} \right)}^{2}}+{{\left( \frac{1}{S-y} \right)}^{2}}+{{\left( \frac{1}{S-z} \right)}^{2}} \right)\ge3 {{S}^{2}}{{\left( \frac{1}{S-\frac{S}{3}} \right)}^{2}}=\frac{27}{4}$, so we must have equality in Jensen, thus $x=y=z$, etc.

Nice solution @above!!! To be honest, when I tried to solve it without Nesbitt 's, I took x+y+z=a but I never imagined that I could use Jensen's to continue!

TuZo wrote: Alternative solution: Because the fractions there are homogenous, we can suppose that $x+y+z=S=constant.$ So we hace ${{S}^{2}}\left( {{\left( \frac{1}{S-x} \right)}^{2}}+{{\left( \frac{1}{S-y} \right)}^{2}}+{{\left( \frac{1}{S-z} \right)}^{2}} \right)=\frac{27}{4}$. But the function $f(t)={{\left( \frac{1}{S-t} \right)}^{2}}$, it is convex, so we have Jensen: ${{S}^{2}}\left( {{\left( \frac{1}{S-x} \right)}^{2}}+{{\left( \frac{1}{S-y} \right)}^{2}}+{{\left( \frac{1}{S-z} \right)}^{2}} \right)\ge {{S}^{2}}{{\left( \frac{1}{S-\frac{S}{3}} \right)}^{2}}=\frac{27}{4}$, so we must have equality in Jensen, thus $x=y=z$, etc. Is typo ? Should be $3{{S}^{2}}{{\left( \frac{1}{S-\frac{S}{3}} \right)}^{2}}$

Yes, it is Only a typo. Thank you.