8/10　今日考えたこと

$$\begin{array}{rl}\sum _{cyc}{\displaystyle \frac{a}{a+nb}}& =\sum _{cyc}{\displaystyle \frac{a^2}{a^2+nb}}\\ & \ge {\displaystyle \frac{(a+b+c{)}^2}{a^2+b^2+c^2+n(ab+bc+ca)}}\end{array}$$
$$\begin{array}{rl}& ⟺{\displaystyle \frac{(a+b+c{)}^2}{a^2+b^2+c^2+n(ab+bc+ca)}}\ge {\displaystyle \frac{3}{n+1}}\\ & ⟺(n+1)(a+b+c{)}^2\ge 3(a^2+b^2+c^2+n(ab+bc+ca))\\ & ⟺(n-2)(a^2+b^2+c^2)\ge (n-2)(ab+bc+ca)\\ & ⟺(a^2+b^2+c^2)\ge (ab+bc+ca)\end{array}$$

$$\begin{array}{rl}\sum _{cyc}{\displaystyle \frac{a}{na+b}}& ={\displaystyle \frac{3}{n}}-{\displaystyle \frac{1}{n}}\sum _{cyc}{\displaystyle \frac{b}{na+b}}\le {\displaystyle \frac{3}{n}}-{\displaystyle \frac{1}{n}}({\displaystyle \frac{3}{n+1}})={\displaystyle \frac{3}{n+1}}\end{array}$$

簡単な式変形から
$$\begin{array}{r}(n-2m)(a^2+b^2+c^2)\ge (n-2m)(ab+bc+ca)\end{array}$$
が成り立つから
$$n-2m>0⟺n\ge 2m+1$$
が条件