Show a function is bounded below by $\sqrt{a}$

Question

I have to show a function $$f(x)= \frac{1}{2}\left(\frac{a}{x}+x\right),$$ where $x>0$, is bounded from below by $\sqrt a$.

The first thing I think of is: if a converges, then it is bounded. So my question: is looking if the limit of $f(x)$ from left and right is $\sqrt a$ the right way to go?

Why would you look for a limit (in this case) as $x\to\infty$ if you want to find a lower bound? — vitamin d, Feb 26 '21 at 20:28
Hint: Show that $x = \sqrt{a}$ is a global minimizer for $f(x)$ when $x > 0$. — DMcMor, Feb 26 '21 at 20:30
I presume you mean for $x>0$, because obviously this is not the case for $x<0$. — , Feb 26 '21 at 20:35

score 1 · Answer 1 · answered Feb 27 '21 at 03:54

The issue with looking at the limits is that it will only tell you things (by definition) at the two extreme points. It won't by itself tell you what the infimum is; just that there is one. You'll have to do extra work afterwards to prove that $\sqrt{a}$ is the infimum (here, minimum).

Here is a simple way to show the statement. By the AM-GM inequality, for every $x>0$ $$ f(x) = \frac{\frac{a}{x}+x}{2} \geq \sqrt{\frac{a}{x}\cdot x} = \sqrt{a}\,. $$

score 0 · Answer 2 · answered Feb 26 '21 at 20:34

0

hint

$ f $ is an odd function.

Assume $ x\ne 0 $, then

$$2x(f(x)-\sqrt{a})=(\sqrt{a})^2+x^2-2x\sqrt{a}$$

$$=(\sqrt{a}-x)^2$$

answered Feb 26 '21 at 20:34

hamam_Abdallah

62,951

score 0 · Answer 3 · answered Feb 27 '21 at 03:50

0

Clearly, $f(x)$ goes to infinity when $\,x\to 0^+$ or $x\to\infty.$ Since $f'(x)=0$ iff $x=\sqrt a, \ \ f(\sqrt a)$ is the lower bound.

answered Feb 27 '21 at 03:50

David

21

Show a function is bounded below by $\sqrt{a}$

3 Answers3