Algebric inequality

Question

I am wondering for an algebric proofs (without calculus, lagrangian etc. because I didn't study yet) for: Let $x,y,z$ positive numbers such that $x+y+z=3$ and $n\in[0,4]$. Show that $$ \frac{1}{n+(x+y)^3}+\frac{1}{n+(y+z)^3}+\frac{1}{n+(z+x)^3}\ge\frac{3}{n+8}. $$ I obteined a proof for $n=0$. We have to prove $$\frac{1}{(x+y)^3}+\frac{1}{(y+z)^3}+\frac{1}{(z+x)^3}\ge\frac{3}{8}. $$ Applying AM-GM twice, it follows $$ \begin{align*} \frac{1}{\left( x+y\right) ^{3}}+\frac{1}{\left( y+z\right) ^{3}}+\frac {1}{\left( z+x\right) ^{3}} & =\frac{1}{\left( 3-z\right) ^{3}}+\frac {1}{\left( 3-x\right) ^{3}}+\frac{1}{\left( 3-y\right) ^{3}}\\ & \geq\frac{3}{\sqrt[3]{\left[ \left( 3-x\right) \left( 3-y\right) \left( 3-z\right) \right] ^{3}}}\\ & =\frac{3}{\left( 3-x\right) \left( 3-y\right) \left( 3-z\right) }\\ & \geq\frac{3}{8},% \end{align*}%$$ because $$ \sqrt[3]{\left( 3-x\right) \left( 3-y\right) \left( 3-z\right) }% \leq\frac{9-x-y-z}{3}=2\Rightarrow\frac{1}{8}\leq\frac{1}{\left( 3-x\right) \left( 3-y\right) \left( 3-z\right) }.% $$ For $n=4$ I did not obtein anything.