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

Question

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 $x\in(0,4)$ and increasing for $x\in[4,\infty)$.

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 $x\in(0,\delta_K)$ 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 $t\geq 2 \ln t$; the point is that I'm not sure if this is valid. I think it is valid because for $x\in(0,\infty)$ the map $x\mapsto t^2$ 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.

Answer 1

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.

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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

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

for $x>0$

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

$$\begin{align} f(t+h)-f(t)&=\frac h2-\log\left(1+\frac ht\right)\\\\ &\ge \frac h2-\frac ht\\\\ &\ge 0 \end{align}$$

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

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

for $t\ge 2$.

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

$$\log(x)\le \sqrt x$$

for $x\ge 4$.

We also have from $(1)$, that $\log(x)\le 2(\sqrt x-1)$. When $x\le 4$, 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!

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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 $x>0$.

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

"So for $x\in(0,\delta_K)$ 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 $\lim_{x\to 0}D(x)=\lim_{x\to\infty}D(x)=\infty$, $D(x)>0$. And we are done.

Answer 2

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

Answer 3

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

Answer 4

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 $x \in (0,\infty)$.