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 $0<x<delta_K$ 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

$$log(t)<t/2 tag 3$$

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 $log(x)<x^alpha$ 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 $x^alpha<x^{alpha+varepsilon}$ 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 $f(t) = e^t - t^2 implies f’(t) = e^t - 2t > 1+ t+ frac{t^2}{2} - 2t = frac{1}{2} + frac{(t-1)^2}{2} > 0 implies f(t) > f(0) = 1 > 0 implies e^t > t^2$ .

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