# Proving that $\ln^2(x) \le Cx$ for some $C >0$ ($x\ge1$)

My proof uses some calculus, and I was wondering and there are any other ways (namely a more elementary way).

Let (for $x\ge1$)$$g(x) = \frac{\ln^2x}{x}$$

Then taking the derivative: $$g'(x) = \frac{2\ln x - \ln^2x}{x^2}$$

We spot that this is positive from $0$ to $e^2$ and decreasing afterwards. Hence $g$ reaches a maximum at $x = e^2$ (and since we start at $x=1$, $g$ is always positive). We can immediately conclude that for any $x\ge1$: $$\ln^2x \le e^2x$$ which proves our claim.

• Can we assume you know $\ln t \leq C't$ for some $C'$? – Clement C. May 22 '18 at 22:34
• @ClementC. Yes we can – Quantaliinuxite May 22 '18 at 22:34

## 2 Answers

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

PRIMER:

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

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

for $x>0$

Let $f(x)=\frac x2-\log(x)$. Then, using $(1)$ we find for $h>0$ that

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

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

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

for $x\ge 2$.

Now, replacing $x$ with $\sqrt{x}$ 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 $\log(x)\le \sqrt x$. Squaring, we find

$$\log^2(x)\le x$$

for $x\ge 1$.

As mentioned in a comment, you know (and thus we can start from there) that there exists some $C'>0$ such that $$\forall t\geq 1,\qquad \ln t \leq C' t\,. \tag{1}$$

Then use the fact that $\ln x = 2\cdot\frac{1}{2}\ln x = 2\ln \sqrt{x}$ to write, since $t \stackrel{\rm def}{=}\sqrt{x}\geq 1$ for $x\geq 1$, $$\forall x\geq 1, \qquad \ln\sqrt{x} \leq C'\sqrt{x}$$ which (everything is positive) is equivalent to $$\forall x\geq 1, \ln^2 \sqrt{x} \leq C'^2 x$$ i.e. $$\forall x\geq 1,\qquad \ln^2 x \leq 4C'^2 x\,. \tag{2}$$ That is, you can set $C\stackrel{\rm def}{=} 4C'^2$.

• Hello my friend! Hope all is well. I just posted a pre-calculus solution in which I show that $\log^2(x)\le x$ for $x\ge 1$. – Mark Viola May 22 '18 at 22:55
• @MarkViola I saw, and updated :) Leaving mine as a way to "bootstrap" a standard result to get this variant. (I like your pre-calculus solutions, in general, but always feel that reducing a new question to a standard result in a few lines is a useful way to go as well) – Clement C. May 22 '18 at 22:57
• Indeed. And (+1) – Mark Viola May 22 '18 at 22:58