解いた不等式まとめ（順次更新中）

$${\displaystyle \frac{a}{b+\sqrt[4]{a}{b}^3}}+{\displaystyle \frac{b}{c+\sqrt[4]{b}{c}^3}}+{\displaystyle \frac{c}{a+\sqrt[4]{c}{a}^3}}\ge \sum _{cyc}{\displaystyle \frac{a}{b+{\displaystyle \frac{1}{4}}a+{\displaystyle \frac{3}{4}}b}}=\sum _{cyc}{\displaystyle \frac{4a}{a+7b}}\ge {\displaystyle \frac{3}{2}}$$
$$⟺\sum _{cyc}{\displaystyle \frac{a}{a+7b}}\ge {\displaystyle \frac{3}{8}}$$
$$⟺\sum _{cyc}{\displaystyle \frac{a^2}{a^2+7ab}}\ge {\displaystyle \frac{a^2+b^2+c^2}{a^2+b^2+c^2+7(ab+bc+ca)}}\ge {\displaystyle \frac{3}{8}}$$
$$⟺8(a^2+b^2+c^2)\ge 8(ab+bc+ca)$$
Muirheadの不等式より明らか

$\mathrm{\Delta }=(三角形の面積)={\displaystyle \frac{abc}{4R}}, R=(外接円の半径)$とする
$$(a+b+c)(3-2a)(3-2b)(3-2c)\le 3a^2{b}^2{c}^2$$
$$⟺16{\mathrm{\Delta }}^2\le 3a^2{b}^2{c}^2$$
$$⟺R^2\ge {\displaystyle \frac{1}{3}}$$
$$⟺\frac{a+b+c}{2(\sin A+\sin B+\sin C)}\ge {\displaystyle \frac{1}{\sqrt{3}}}$$
$$⟺\sin A+\sin B+\sin C\le \frac{3\sqrt{3}}{2}$$
Jensenの不等式より明らか