Why Is ln(1+x) Greater Than x/(2+x) for x > 0?

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SUMMARY

The inequality ln(1+x) > x/(2+x) holds true for all x > 0. This is established by defining the function f(x) = ln(1+x) - x/(2+x) and demonstrating that its derivative, f'(x) = (x^2 + 2x + 2)/((1+x)(2+x)^2), is positive for all x > 0. Consequently, f(x) is strictly increasing in the interval (0, +∞). Additionally, the limit as x approaches 0 from the right confirms that f(0) = 0, reinforcing the validity of the inequality.

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How to prove that for x>0
\ln(1+x) > \frac{x}{2+x} is true?
 
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Lisa91 said:
How to prove that for x>0
\ln(1+x) > \frac{x}{2+x} is true?

Yes, it is true. One way to prove it: denote, $f(x)=\ln(1+x) - \dfrac{x}{2+x}$, then $f'(x)=\ldots=\dfrac{x^2+2x+2}{(1+x)(2+x)^2}>0$ for all $x>0$. This means that $f$ is strictly increasing in $(0,+\infty)$. On the other hand,

$\displaystyle\lim_{x\to 0^+}f(x)=\displaystyle\lim_{x\to 0^+}\left(x+o(x)-\frac{x}{2+x}\right)=0$.
 

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