$\log(x+c)\leqslant x$? Proof 
I am in need of this important inequality $$\log(x+1)\leqslant x$$.

I understand that $\log(x)\leqslant x$. For $c\in\mathbb{R}$.  However is it true that $\log(x+c)\leqslant x$?
It is hard to accept because it seems like $c$ cannot be arbitrary.
I have tried to prove this inequality:
$\log(x+c)\leqslant x\iff x+c\leqslant e^x$
It is true that $f(x)=x$ grows much faster than $g(x)=\log(x+c)$, since the $\frac{df(x)}{dx}=1\geqslant \frac{1}{x+c}=\frac{dg(x)}{dx}$ 
Question:
Is the derivative argument enough to prove the more general inequality  $\log(x+c)\leqslant x$?
Thanks in advance!
 A: Hint: write $$x-\log(x+1)\geq 0$$ and define $$f(x)=x-\log(x+1)$$ and use calculus. And $$f'(x)=\frac{x+1-x}{x(x+1)}$$
A: The derivative argument works. For your specific inequality, note that the derivative argument can be written in terms of integrals:
$$y-1=\int_1^y\frac11{\rm~d}t\ge\int_1^y\frac1t{\rm~d}t=\ln(y)$$
where $y=x+1>1$.
For $c<1$ it follows trivially from above. For $c>1,y>0$, note that
\begin{align}y+c&\ge y+\ln(c)+1\\&>y+\ln(c)+\ln(2)\\&=y+\ln(2c)\\&=\ln(2c)+\int_{2c}^{y+2c}1{\rm~d}t\\&\ge\ln(2c)+\int_{2c}^{y+2c}\frac1t{\rm~d}t\\&=\ln(y+2c)\end{align}
Let $y+c=x$ and you end up with
$$x\ge\ln(x+c)\quad\forall x>c$$
A: "However is it true that log(x+c)⩽x?"
Let $x =1$ and $c = 500,000,000$ then is $\log 500,000,001 \le 1$?
It is true that if $\log (x+c) \le x\iff x+c \le e^x$ and it is true that $\frac {d(x+c)}{dx} = 1 \le \frac{d e^x}{dx} = e^x$ if $x > 0$.  So, for positive $x$ we have that $e^x$ increases faster than $x + c$.
But it's not enough that a function increases (has a greater derivative) faster than another function to ensure it is always greater.  The function must also have a greater initial value.
If $0 + c > e^0 = 1$ then there will but some $(b,d)$ where $x \in (b,d)$ will imply $x + c > e^x$.  ($b,d$ will be the two solutions to $e^x - x = c$.  It's easy to convince ourselves that if $c > 1$ then there are exactly two such points where $b < 0 < d$.  And that if $x \ge d$ then $\log (x + c) \le x$.)
A: Hint: The claim is equivalent to $$ e^x\ge 1+x$$
