Question

If a, b, c are positive real number, then minimum value of $\frac{a^3}{4b}+\frac{b}{8c^2}+\frac{1+c}{2a}$, is

• A. $\frac{3}{4}$
• B. $\frac{2}{5}$
• C. $\frac{5}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{5}{4}$

$\frac{\frac{a^3}{4b}+\frac{b}{8c^2}+\frac{1}{2a}+\frac{c}{4a}+\frac{c}{4a}}{5}\ge {(\frac{a^3}{4b}.\frac{b}{8c^2}.\frac{1}{2a}.\frac{c}{4a}.\frac{c}{4a})}^{\frac{1}{5}}$
$\frac{a^3}{4b}+\frac{b}{8c^2}+\frac{1+c}{2a}\ge \frac{5}{4}$