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]

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?