I want to show that

$$\log(2+x) - \log(x) \lt \frac{2}{x}$$ $\forall x \in \mathbb{R^+}$

I know I need to apply the Mean Value Theorem to find an upper bound of the function to the left and show that it is smaller than $\frac{2}{x}$, but I can't find the correct upper bound. I've tried multiple variations of the inequality. My teacher also said that I only needed to check in the interval from 0 to 2, but I'm not sure why.

  • $\begingroup$ $\exists c\in(x,x+2)\Rightarrow (log)'(c)=\frac{log(2+x)-log(x)}{2+x-x}$ $\endgroup$ – Mateus Rocha Jul 11 '18 at 0:41
  • $\begingroup$ @MateusRocha why are you applying to $log(x)$? It should be to $log(2+x)-log(x)$ $\endgroup$ – chilliefiber Jul 11 '18 at 0:45
  • $\begingroup$ using what I've commented, $\frac{1}{c}=\frac{log(2+x)-log(x)}{2}\Rightarrow \frac{2}{c}=log(2+x)-log(x)$. But $c>x$, so $\frac{2}{c}<\frac{2}{x}$ $\endgroup$ – Mateus Rocha Jul 11 '18 at 0:48
  • $\begingroup$ Essentially the same problem as in math.stackexchange.com/questions/2067473/…. $\endgroup$ – Martin R Jul 11 '18 at 7:48

For every real $\alpha>0$, let $f(x)=\ln(x+\alpha)$. Now with Mean Value Theorem in $[0,2]$ : $$f(2)-f(0)=f'(c)(2-0)$$ for a $c\in(0,2)$. This gives us $\ln(2+\alpha)-\ln\alpha<\dfrac{2}{\alpha}$.

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