How do I find the lower bound of $a_n:=n+\frac{100}{n}$ without inserting values?

Question

How do I find the lower bound of $a_n:=n+\frac{100}{n}$ without inserting values?

I already found out, that the sequence has no upper bound, because $\lim\limits_{n\to \infty}a_n\to\infty$. However, I'm not sure how to go on with the lower bound.

Answer 1

From the inequality between arithmetic and geometric mean: $$ \frac 12 a_n = \frac 12 \left(n + \frac{100}{n} \right) \ge \sqrt{n \cdot \frac{100}{n}} = 10 \, . $$ Equality holds if (and only if) $n = \frac{100}{n}$, that is for $n=10$.

Answer 2

Hint: This is essentially asking how small the sum of two (positive) numbers can be, given that they multiply to $100$. What does intuition tell you about when the sum of two numbers is minimized when something like their product should be fixed? Can you answer the question when the product is $1$ instead of $100$ and then extrapolate?

On the other hand, if you're familiar with inequalities, something like AM-GM could help.

Answer 3

$a_n= n+100/n=$

$ (\sqrt{n})^2 + (\sqrt{100/n})^2 =$

$(\sqrt{n}-\sqrt{100/n})^2 + 2\sqrt{n}\sqrt{100/n} $

$\ge 2\sqrt{n(100/n)} = 20;$

Equality for $(n)^{1/2}=(100/n)^{1/2}$, i.e

$n=100/n$.