# $\log(x+c)\leqslant x$? 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 $\log(x+c)\leqslant x$?

• If $x = 3; c = 3,569$ is $\log(3 + 3,569) \le 3$? – fleablood Apr 8 '18 at 16:14
• It is true that it grows faster but it is not true that it starts equal. If $x_0 + c > e^{x_0}$ then is doesn't matter then $f'(x) \ge g'(x)$. – fleablood Apr 8 '18 at 16:17
• "Is the derivative argument enough to prove the more general inequality log(x+c)⩽x?" Only if $x \ge d$ where $d$ is a positive solution to $e^x - x = c$ – fleablood Apr 8 '18 at 16:33

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

$$x\ge\ln(x+c)\quad\forall x>c$$

Hint: write $$x-\log(x+1)\geq 0$$ and define $$f(x)=x-\log(x+1)$$ and use calculus. And $$f'(x)=\frac{x+1-x}{x(x+1)}$$

"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 $\log (x + c) \le x$.)

• This is the only answer posted as of now that actually addresses the question at the beginning of your solution. (+1) May I suggest your adding something about the correct inequality $\log(1+x)\le x$ for all $x$? – Mark Viola Apr 8 '18 at 16:39
• $f\gg g$ often denotes $\exists N\forall x>N[f(x)>g(x)]$ i.e. the inequality holds for sufficiently large $x$. – Simply Beautiful Art Apr 9 '18 at 1:46

Hint: The claim is equivalent to $$e^x\ge 1+x$$