How to prove ln(x) < sqrt(x) for all x>0

grossgermany

Hi,

I know I can use a graphical calculator to easily show that
How to prove ln(x) < sqrt(x) for all x>0

But I wonder if there is a rigorous way to demonstrate this.

fourier jr

what do you call that? proof by calculator graphic? I would consider the function $f(x) = lnx - \sqrt{x}$ (or is it the other way around? well you get the idea) & use calculus to show that the resulting function is monotone increasing on that interval. off the top of my head that's probably the best way to do it.

grossgermany

Is there a non-graphical way of doing it?

Vardd

Hi Grossgermany,

To answer the question, I'm assuming you have a rough idea of Calculus.

We know that the derivative of a formula, tells us the rate of change of the formula. We also that both formulas are always positive.

(Sorry no Latex I tried to make it works, but i failed. )

So let's take the derivative of ln(x) d/dx (ln(x)) = 1/x
Then of sqrt(x) d/dx (sqrt(x)) = 1/(2sqrt(x))

As x goes to infinity, 1/2 becomes useless, and for simplicity sake let's remove it.

1/x < 1/(2sqrt(x)) --> 1/x < 1/sqrt(x)

Then we can clearly see that as x increase to infinity, the rate of change of sqrt(x) is greater then ln(x).

I've just though about it while reading your question, but I am fairly sure of my answer.

Vardd

JonF

Showing the derivative of x^1/2 – ln(x) is positive for all x>0 is fairly easy.

JonF

Hi Grossgermany,

To answer the question, I'm assuming you have a rough idea of Calculus.

We know that the derivative of a formula, tells us the rate of change of the formula. We also that both formulas are always positive.

(Sorry no Latex I tried to make it works, but i failed. )

So let's take the derivative of ln(x) d/dx (ln(x)) = 1/x
Then of sqrt(x) d/dx (sqrt(x)) = 1/(2sqrt(x))

As x goes to infinity, 1/2 becomes useless, and for simplicity sake let's remove it.

1/x < 1/(2sqrt(x)) --> 1/x < 1/sqrt(x)

Then we can clearly see that as x increase to infinity, the rate of change of sqrt(x) is greater then ln(x).

I've just though about it while reading your question, but I am fairly sure of my answer.

Vardd
You can’t just remove the ½, and your method would only show for some value of N, for x>N would ln(x) < x^1/2

Anonymous217

Showing the derivative of x^1/2 – ln(x) is positive for all x>0 is fairly easy.
This is the way to do it. You also need to show that it's a positive value from the beginning though (lim x approaches 0 of f(x) = +) since increasing wouldn't really prove anything by itself (ex: any function where f(x) = - where x< some positive number k, but f(x) increasing from x>0).

grossgermany

This is the way to do it. You also need to show that it's a positive value from the beginning though (lim x approaches 0 of f(x) = +) since increasing wouldn't really prove anything by itself (ex: any function where f(x) = - where x< some positive number k, but f(x) increasing from x>0).
Except the statement is not true
0.5*1/sqrt(x)-1/x is not always positive for x>0.

JonF

Except the statement is not true
is not always positive for x>0.
good catch, so where would the min of 0.5*1/sqrt(x)-1/x be at?

fourier jr

whoops, it's not monotone like I said but the minimum is positive, & that will still do it

alomari2010

consider the function $${f(x) = e^{\sqrt{x}} - x,}$$ which is >0
so that $$f^{\prime}$$ >0 for all x and therefore
$$e^{\sqrt{x}} - x > 0$$ , and thus $$\sqrt{x} > \ln{x}$$

CRGreathouse

Homework Helper
whoops, it's not monotone like I said but the minimum is positive, & that will still do it
consider the function $${f(x) = e^{\sqrt{x}} - x,}$$ which is >0
so that $$f^{\prime}$$ >0 for all x and therefore
$$e^{\sqrt{x}} - x > 0$$ , and thus $$\sqrt{x} > \ln{x}$$
How do you prove that it's positive?

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