1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ If $x = 3; c = 3,569$ is $\log(3 + 3,569) \le 3$? $\endgroup$ – fleablood Apr 8 '18 at 16:14
  • $\begingroup$ 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 $f'(x) \ge g'(x)$. $\endgroup$ – fleablood Apr 8 '18 at 16:17
  • $\begingroup$ "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 $e^x - x = c$ $\endgroup$ – fleablood Apr 8 '18 at 16:33
2
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 $x$? $\endgroup$ – Mark Viola Apr 8 '18 at 16:39
  • $\begingroup$ $f\gg g$ often denotes $\exists N\forall x>N[f(x)>g(x)]$ i.e. the inequality holds for sufficiently large $x$. $\endgroup$ – Simply Beautiful Art Apr 9 '18 at 1:46
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.