Question

If a $>0,$ then the minimum value of sum of $\frac{1}{a},1,a^2,a^3,\frac{1}{a^4}$ is equal to

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

Sum of $\frac{1}{a},1,a^2,a^3,\frac{1}{a^4}$ is $1+a^2+a^3+\frac{1}{a}+\frac{1}{a^4}$

$AM\ge GM$

$⟹\frac{1+a^2+a^3+\frac{1}{a}+\frac{1}{a^4}}{5}\ge \sqrt[5]{5\left(1\right)\left(a^2\right)\left(a^3\right)\left(\frac{1}{a}\right)\left(\frac{1}{a^4}\right)}$

$1+a^2+a^3+\frac{1}{a}+\frac{1}{a^4}\ge 5\sqrt[5]{51}$

$1+a^2+a^3+\frac{1}{a}+\frac{1}{a^4}\ge 5$

The minimum value of $1+a^2+a^3+\frac{1}{a}+\frac{1}{a^4}$ is $5$