$\text{Show that }\log(1+x) \le x , \quad \forall x \in [0,\infty)$ Show that 
$$\log(1+x) \le x , \quad \forall x \in [0,\infty)$$

So I have come up with two ways to solve this question and I was wondering which way is more accurate?
Proof 
$\boldsymbol{\text{Way 1}} \\ \text{Let } f(x) = \log(1+x) -x \le 0 \\ \text{Notice f(0) = 0 } \\ f'(x) \le 0 \quad x \in [0,\infty) \\ f(x) \text{ is decreasing } \\ \text{ Now } f'(x) = \frac{-x}{1+x}\le 0 \text{ True.} $ 
$\boldsymbol{\text{Way 2}}\\ \text{Let} f(x) = x- \log(1+x) \ge 0 \\ f(0) =0 \\ \text{We only NTS } \\ f'(x) \ge 0 \\ \text{f(x) is increasing}  \\ f'(x) =1-\frac{1}{1+x} = \frac{x}{1+x} \ge 0 \text{ True } $
 A: Let $g(x)=\log(1+x)$. Since $$g''(x)=-\frac{1}{(1+x)^2}<0$$
$g$ is a concave function and $y=x$ is their tangent  line at the point $(0,0)$.
A: If you define
$\log(x)
=\int_1^x \dfrac{dt}{t}
$,
then
$\begin{array}\\
\log(1+x)
&=\int_1^{1+x} \dfrac{dt}{t}\\
&=\int_0^{x} \dfrac{dt}{1+t}\\
&\lt\int_0^{x} dt\\
&=x\\
\end{array}
$
A: The proofs have all the right ingredients (though it's not really two different ways, it's the same method just with a slightly different definition of $f$). What you can improve is the presentation of the proof. Try to write it up in a more structured and logical way and add some words. Right now you are mixing up the definition of the function with the inequality you want to show and then mixing up referencing the theorem you use with showing that your function satisfy it which is a bit confusing.
Here is a simple example of how one could rewrite your proof:

Define the function $f(x) = x - \log(1+x)$. We see that $f'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x}$ so $f'(x) \geq 0$ for $x\geq 0$. Since the derivative is positive $f$ is increasing on $[0,\infty)$ and we therefore have 
  $$f(x) \geq f(0) = 0  \implies x\geq \log(1+x)~~~\text{for}~~~x\geq 0$$

If you want to be more detailed you could for example also prove the theorem you use:

If $f'(x) \geq 0$ then from the mean value theorem we have $\frac{f(b)-f(a)}{b-a} = f'(c) \geq 0$ and it follows that if $b>a$ then $f(b) \geq f(a)$ so $f$ is increasing.

Or just reference it:

By the mean value theorem $f'(x) \geq 0$ implies that $f$ is increasing and we therefore have...

However this is such a well known statement that it's not needed to prove it unless you want to be very thorough.
A: You can add an exponent on both sides. The result is
$$1+x \leq \mathrm{e}^x$$
Remembering that $\mathrm{e}^x = \sum_{i=0}^\infty \frac{x^i}{i!}$, the inequality holds since the l.h.s. is exactly the first two elements of the exponent Taylor series, and all elements in this series are positive.
A: If $x=0$ then the relation is valid. Let $x>0$. By the mean value theorem we have that
$$
\frac{\log(1+x)-\log 1}{1+x-1}=\frac 1t,\qquad t\in(1,1+x).
$$
Since the right-hand side is less than 1 we get the given relation
$$
\frac{\log(1+x)}{x}<1\quad\Rightarrow\quad \log(1+x)<x,\quad x>0.
$$
