# Prove that $sqrt{x} > ln x$ for all $x>0$ with a study of function

Mathematics Asked by ZaWarudo on September 1, 2020

Prove that for all $$x>0$$ it is $$sqrt{x} > ln x$$.

I’ve defined the function $$D(x)=sqrt{x}-ln x$$, I’ve studied its first derivative and I’ve concluded that $$D$$ is decreasing for $$xin(0,4)$$ and increasing for $$xin[4,infty)$$.

Since $$D(4)>0$$ for monotonicity I can conclude that the inequality is true in the interval $$[4,infty)$$, but now I’m unsure on how study the thing in the interval $$(0,4)$$.

Since $$D(4)>0$$ it is enough to show that $$D$$ is positive in $$(0,4)$$, I would like to use the fact that $$D$$ is decreasing in $$(0,4)$$ but I can’t calculate $$D(0)$$ since $$D$$ is not defined at $$x=0$$; so I would like to use the fact that $$D$$ tends to $$infty$$ as $$x to 0^+$$, by that it follows that for all $$K>0$$ there exists $$delta_K>0$$ such that if $$0 it is $$D(x)>K>0$$.

So for $$xin(0,delta_K)$$ it is $$D(x)>0$$, then $$D$$ is decreasing and since $$D(4)>0$$ it follows that $$D$$ is positive in the interval $$(0,4)$$ as well. Is this correct?

Another question, an alternative approach was the following: since $$x>0$$ I thought about letting $$x=t^2$$, so the statement would be equivalen to $$tgeq 2 ln t$$; the point is that I’m not sure if this is valid. I think it is valid because for $$xin(0,infty)$$ the map $$xmapsto t^2$$ is bijective and so I don’t lose informations, so solving the inequality in $$t$$ the other set where $$t$$ varies (which in this case is the same of the initial set) is equivalent of solving it in the initial set. Is this correct? I mean, if I do these kind of substitutions, I have to check that they are invertible? Thanks.

Herein, we show using non-calculus based tools that $$log(x)le sqrt{x}$$ for all $$x>0$$. We begin with a primer on elementary inequalities for the logarithm function.

PRIMER:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$frac{x-1}{x}le log(x)le x-1tag1$$

for $$x>0$$

Let $$f(t)=frac t2-log(t)$$. Then, using $$(1)$$ we find for $$h>0$$ that

begin{align} f(t+h)-f(t)&=frac h2-logleft(1+frac htright)\\ &ge frac h2-frac ht\\ &ge 0 end{align}

for all $$tge 2$$. So, $$f(t)$$ is monotonically increasing for $$tge 2$$. And since $$f(2)=1-log(2)>0$$ we have

$$log(t)

for $$tge 2$$.

Now, setting $$t= sqrt{x}$$ in $$(3)$$ reveals

$$log(x)le sqrt x$$

for $$xge 4$$.

We also have from $$(1)$$, that $$log(x)le 2(sqrt x-1)$$. When $$xle 4$$, we see that $$2(sqrt x-1)le sqrt x$$.

Hence, for all $$x>0$$, we find that $$log(x)le sqrt x$$ as was to be shown!

NOTE:

It might be of interest that the smallest number $$alpha$$ for which $$log(x) for all $$x>0$$ is $$alpha =1/e$$. For $$alpha=1/e$$, $$log(x)=x^alpha$$ at $$x=e^e$$ where the slopes of curves $$y=log(x)$$ and $$y=x^{1/e}$$ are equal.

This is not a surprising result. For any two smooth functions $$f(x)$$ and $$g(x)$$ for which $$f(x)ge g(x)$$ and $$f(x_0)=g(x_0)$$ for some point $$x_0$$, $$x_0$$ is a local minimum with $$f'(x_0)=g'(x_0)$$.

If $$f(x)=x^alpha$$ and $$g(x)=log(x)$$, then $$x_0^alpha=log(x_0)$$ and $$alpha x_0^{alpha-1}=x_0^{-1}$$ from which we find $$alpha=1/e$$ and $$x_0=e^e$$.

Finally, since $$x^alpha for all $$varepsilon>0$$ and $$x>1$$, we conclude that $$log(x)< x^beta$$ for all $$beta>1/e$$ and $$x>0$$.

EDIT: I want to address the specific question of the OP.

"So for $$xin(0,delta_K)$$ it is $$D(x)>0$$, then $$D$$ is decreasing and since $$D(4)>0$$ it follows that $$D$$ is positive in the interval $$(0,4)$$ as well. Is this correct?"

Yes, the argument is correct. But things are even simpler.

Just note that in the domain of definition, $$x>0$$, of $$D(x)$$, $$D'(4)=0$$. Hence, $$x=4$$ is a local extremum of $$D(x)$$. Moreover, $$D'(x)<0$$ for $$x<4$$ and $$D'(x)>0$$ for $$x>4$$. So, $$x=4$$ is a local minimum.

Inasmuch as $$D(4)>0$$ and $$lim_{xto 0}D(x)=lim_{xtoinfty}D(x)=infty$$, $$D(x)>0$$. And we are done.

Correct answer by Mark Viola on September 1, 2020

I decide it in following way: as $$D(x)=sqrt{x}-ln x$$ is decreasing for $$x in (0,4)$$ and increasing for $$x in [4,infty)$$, then D(4) is minimum. $$D(4)=2-ln 4 approx 2-1.386294361 geqslant 0$$, so $$D(x) geqslant 0$$ for $$x in (0,infty)$$.

Answered by zkutch on September 1, 2020

If $$xin (0,1]$$ then is obvius that $$sqrt{x}>ln{x}$$, Define $$f(x)=frac{sqrt{x}}{ln{x}}$$ With $$xin(1,infty)$$, then $$f(x)>0$$ for all $$x$$, and has minimun in $$x=e^2$$, then: $$f(x)geqfrac{e}{2}>1impliesfrac{sqrt{x}}{ln{x}}>1$$

Answered by AsdrubalBeltran on September 1, 2020

Put $$t = sqrt{x} implies e^t > t^ 2, 0 < t < 2$$. Consider $$f(t) = e^t - t^2 implies f’(t) = e^t - 2t > 1+ t+ frac{t^2}{2} - 2t = frac{1}{2} + frac{(t-1)^2}{2} > 0 implies f(t) > f(0) = 1 > 0 implies e^t > t^2$$ .

Answered by DeepSea on September 1, 2020

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