I have to prove the inequality $\frac{x}{(x+1)} < \ln(1+x)<x$ for $x>0$. I don't even know how to relate Lagrange to this, how can I transform this so I can apply Lagrange?
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Find the minima of the functions $x - \ln(1+x)$ and $\ln(1 + x) - x/(x+1)$. For the first case, we have: $$\frac{d}{dx} ( x - \ln(1+x)) = 1 - \frac{1}{1+x} > 0$$ for x > 0. Hence $x - \ln(1+x)$ is a (strictly) monotone increasing function and therefore $x - \ln(1+x) > 0 - \ln(1+0) = 0$, or equivalently $x > \ln (1+x)$. |
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(1) Define the function $$f(x):=\log(1+x)-\frac{x}{1+x}\Longrightarrow f'(x)=\frac{1}{x+1}-\frac{(x+1)-x}{(x+1)^2}=\frac{x}{(x+1)^2}\ge 0\,\,\,\forall\,x\ge0$$ From the above it follows at once that $\,f\,$ is monotone non-descending in $\,[0,\infty)\,$, so $$\forall\,x\ge 0\;\;,\;\;f(x)\ge f(0)=0$$ and we have the left hand inequality. Try now something similar for the right hand inequality... |
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Hint: Besides to other answers, you could do that via The Mean Value Theorem. Set $f(x)=\ln(x)$ and by using above theorem show that for $0<a<b$ we have $$1-\frac{a}{b}<\ln(b/a)<\frac{b}{a}-1$$. In fact, there is a $0<a<\xi<b, f'(\xi)=\frac{1}{b-a}(f(b)-f(a))$. Now note that if $x>0$, then $x+1>x>0$ and... |
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