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. 
 A: 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 $C\stackrel{\rm def}{=} 4C'^2$.
A: 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 $x\ge 1$.  
