parmenides51 wrote: Let $a$, $b$, $c$ be positive real numbers. Determine the minimum value of the following expression $$ \frac{a^2+2b^2+4c^2}{b(a+2c)}$$ By AM-GM, $$ \frac{a^2+2b^2+4c^2}{b(a+2c)}=\frac{a^2+b^2+b^2+4c^2}{b(a+2c)}\ge\frac{2ab+4bc}{b(a+2c)}=2$$