For n=3 we can rewrite the inequality: $\sum\limits_{cyc}^{{}}{\frac{{{a}_{1}}.{{a}_{2}}}{2.{{a}_{3}}+{{a}_{1}}+{{a}_{2}}}}=\sum\limits_{cyc}^{{}}{\frac{{{a}_{1}}.{{a}_{2}}}{({{a}_{3}}+{{a}_{1}})+({{a}_{3}}+{{a}_{2}})}}\le Cauchy\le \sum\limits_{cyc}{\left[ \frac{{{a}_{1}}.{{a}_{2}}}{4({{a}_{3}}+{{a}_{1}})}+\frac{{{a}_{1}}.{{a}_{2}}}{4({{a}_{3}}+{{a}_{2}})} \right]}=\frac{1}{4}$ If n>3, than \[\sum\limits_{cyc}^{{}}{\frac{{{a}_{1}}.{{a}_{2}}.\cdots {{a}_{n-1}}}{{{a}_{n}}+n-2}}\le A.G.O.\le \sum\limits_{cyc}^{{}}{\frac{{{\left[ \frac{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n-1}}}{n-1} \right]}^{n-1}}}{{{a}_{n}}+n-2}}<\frac{{{\left[ \frac{{{a}_{1}}+{{a}_{2}}+\cdots +{{a}_{n-1}}+{{a}_{n}}}{n-1} \right]}^{n-1}}}{n-2}=\frac{1}{{{\left[ n-1 \right]}^{n-1}}n-2}<\frac{1}{{{\left[ n-1 \right]}^{2}}}\]

In fact, it looks like the stronger inequality holds too: \[ \frac{a_{2}\cdot a_{3}\cdot\dots\cdot a_{n}}{a_{1}+n-2}+\frac{a_{1}\cdot a_{3}\cdot\dots\cdot a_{n}}{a_{2}+n-2}+\dots+\frac{a_{1}\cdot a_{2}\cdot\dots\cdot a_{n-1}}{a_{n}+n-2}\leq\frac{n^{3-n}}{\left(n-1\right)^{2}} \]

me@home wrote: In fact, it looks like the stronger inequality holds too: \[ \frac{a_{2}\cdot a_{3}\cdot\dots\cdot a_{n}}{a_{1}+n-2}+\frac{a_{1}\cdot a_{3}\cdot\dots\cdot a_{n}}{a_{2}+n-2}+\dots+\frac{a_{1}\cdot a_{2}\cdot\dots\cdot a_{n-1}}{a_{n}+n-2}\leq\frac{n^{3-n}}{\left(n-1\right)^{2}} \]

If $n=3$, $LHS=\sum\limits_{cyc}\frac{a_2a_3}{a_1+1}=\sum\limits_{cyc}\frac{a_2+a_3}{(a_1+a_2)+(a_1+a_3)}\leqslant\sum\limits_{cyc}\frac {a_2a_3}4\left(\frac{1}{a_1+a_2}+\frac{1}{a_1+a_3}\right)=\frac 14$. If $n\geq 4$, $LHS\leqslant \sum\limits_{k=1}^n\dfrac{\left(\frac 1{n-1}\sum\limits_{1\leq i\neq k\leq n}a_i\right)^{n-1}}{a_1+n-2}=\sum\limits_{k=1}^n\frac{1-a_1}{(n-1)^{n-1}(a_1+n-2)}\leqslant \frac 1{(n-1)^{n-1}}\sum\limits_{k=1}^n\frac 1{n-2}\leqslant RHS$.$\blacksquare$