parmenides51 wrote: Let $a,b,c>0$ and $a+b+c=6$ . Prove that $$ \frac{1}{a^2b+16}+\frac{1}{b^2c+16}+\frac{1}{c^2a+16} \ge \frac{1}{8}.$$ This solution use only AM-GM inequality . \begin{align*}\sum_{cyc}\frac{1}{a^2b+16} &= \frac{1}{16}\left(3-\sum_{cyc}\frac{a^2b}{a^2b+16}\right) \\&=\frac{1}{16}\left(3-\sum_{cyc}\frac{a^2b}{a^2b+8+8}\right) \\&\ge\frac{1}{16}\left(3-\sum_{cyc}\frac{a^2b}{12\sqrt[3]{a^2b}}\right)\\&=\frac{1}{16}\left(3-\sum_{cyc}\frac{\sqrt[3]{a^4b^2}}{12}\right)\\&\ge\frac{1}{16}\left(3-\sum_{cyc}\frac{a^2+ab+ab}{36}\right)\\&=\frac{1}{16}\left(3-\frac{(a+b+c)^2}{36}\right)\\&=\frac{1}{8}. \end{align*}