ISI2017-MMA-7

Question

Let $n \geq 3$ be an integer. Then the statement $(n!)^{1/n} \leq \dfrac{n+1}{2}$ is

• A. true for every $n \geq 3$
• B. true if and only if $n \geq 5$
• C. not true for $n \geq 10$
• D. true for even integers $n \geq 6$, not true for odd $n \geq 5$

Answer 1

By AM-GM inequality, We have $$\sqrt[n]{1\cdot2\cdot3\cdot...\cdot n}\le\frac{1+2+3+...+n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}{2}$$

$\Rightarrow ( n!)^{\frac{1}{n}}\leq \frac{n+1}{2}$

Hence, Option $(A)$ is correct.