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

Question

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!

Answer 1

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

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

Answer 2

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

Answer 3

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

Answer 4

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