$a,b,c \in \mathbb{R}^+, n\geq3, n\in \mathbb{N}$
$$\dfrac{a}{a+nb}+\dfrac{b}{b+nc}+\dfrac{c}{c+na} \geq \dfrac{3}{n+1}$$
\begin{align*}
\sum_{cyc}\dfrac{a}{a+nb}&=\sum_{cyc}\dfrac{a^2}{a^2+nb}\\
&\geq \dfrac{(a+b+c)^2}{a^2+b^2+c^2+n(ab+bc+ca)}\\
\end{align*}
\begin{align*}
&\iff \dfrac{(a+b+c)^2}{a^2+b^2+c^2+n(ab+bc+ca)} \geq \dfrac{3}{n+1}\\
&\iff (n+1)(a+b+c)^2 \geq 3(a^2+b^2+c^2+n(ab+bc+ca)) \\
&\iff (n-2)(a^2+b^2+c^2) \geq (n-2)(ab+bc+ca) \\
&\iff (a^2+b^2+c^2)\geq (ab+bc+ca)
\end{align*}
$$\dfrac{a}{na+b}+\dfrac{b}{nb+c}+\dfrac{c}{nc+a} \leq \dfrac{3}{n+1}$$
\begin{align*} \sum_{cyc}\dfrac{a}{na+b}&=\dfrac{3}{n}-\dfrac{1}{n}\sum_{cyc}\dfrac{b}{na+b} \leq \dfrac{3}{n}-\dfrac{1}{n}(\dfrac{3}{n+1}) =\dfrac{3}{n+1} \\ \end{align*}
$$\dfrac{a}{ma+nb}+\dfrac{b}{mb+nc}+\dfrac{c}{mc+na} \geq \dfrac{3}{m+n}$$
簡単な式変形から
\begin{align*}
(n-2m)(a^2+b^2+c^2) \geq (n-2m)(ab+bc+ca)
\end{align*}
が成り立つから
$$n-2m>0 \iff n \geq 2m+1$$
が条件