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Proof Verification: $\frac{x}{x+1} \leq \ln(x+1)$ for $x > -1$ with equality iff $x=0$

My Attempt:

We know that $\frac{1}{(x+1)^2} \leq \frac{1}{x+1}$, for $x > 0. $At $x=0 $ they are the same.

$$\int\frac{1}{(x+1)^2}dx \leq \int\frac{1}{x+1}dx$$ $$\frac{-1}{x+1} +C_1 \leq \ln(x+1) + C_2$$

We conclude that $C_1 =1, C_2 =0$ f0r the $x=0$ case from the original equality to hold.

$$\frac{-1}{x+1}+\frac{x+1}{x+1}-1 +1 \leq \ln(x+1)$$ $$\frac{x}{x+1} \leq \ln(x+1)$$

Is this approach right?

A. Radek Martinez
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  • It looks good to me. – callculus42 Nov 25 '20 at 00:47
  • it looks good to me as well, but I think you might be able to get away with just saying: Look they are the same at $x=0$, and $ln(x+1)$ grows faster (take derivatives) and you are done. – Samael Manasseh Nov 25 '20 at 00:54
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    The first line "$\frac{1}{(x+1)^2} \leq \frac{1}{x+1}$" is false for $-1 < x < 0$ (try plugging in $x = -\frac{1}2$), so this approach will need to consider that. For an alternative approach, take a look at https://math.stackexchange.com/questions/324345/intuition-behind-logarithm-inequality-1-frac1x-leq-log-x-leq-x-1 – Seokbin Lee Nov 25 '20 at 00:59
  • Is there a quick fix for that? Cause I need to show its true for $x>-1$ not $x>0$ – A. Radek Martinez Nov 25 '20 at 01:02
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    Why not take the derivative of $\ln(x+1)-\frac{x}{x+1} $? It's as simple as $x/(x+1)^2$. – Michael Hoppe Nov 25 '20 at 10:37

1 Answers1

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Your answer is ok. But it is better to integrate $$ \frac{1}{(t+1)^2} \leq \frac{1}{t+1}$$ from $0$ to $x$; namely $$ \int_0^x\frac{1}{(t+1)^2}dt \leq \int_0^x\frac{1}{t+1}dt. $$ Then you will get the desired inequality soon instead of concluding $C_1=C_2=0$.

xpaul
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