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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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Why do you think this has anything to do with "Lagrange"? What do you mean by "Lagrange" anyway? –  Chris Eagle Feb 3 '13 at 11:20
    
Well my intuition tells me that has smth to do with it..because this is the chapter "Mean value theorems" but maybe im wrong.. –  superbass Feb 3 '13 at 11:22
    
Apply Lagrange (the Mean Value Theorem) to the function $f(a)=\ln(1+a)$ on the interval $[0,x]$. This gives a $0<c<x$ such that $\ln(1+x)={x\over 1+c}$. Now note, since $0<c<x$, the left hand side of the preceding is strictly bounded above by $x$ and below by ${x\over 1+x}$. –  David Mitra Feb 3 '13 at 13:05
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3 Answers

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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Nice answer, Don. + –  B. S. Feb 3 '13 at 11:41
    
You need to be a bit more carfull, to conclude the strict inequality $f(x) > f(0)$. In fact this can be done by the mean-value theorem: $f(x) - f(0) = x f'(\xi)$ for some $\xi \in (0,x)$ and there your calculation shows $f'(\xi) > 0$. –  Sam Feb 3 '13 at 11:54
    
It's simpler, imo, as it happens that $\,f'>0\,\,\,\forall x>0\,$ –  DonAntonio Feb 3 '13 at 11:56
    
I agree. I gave this argument with the mean-value theorem, since superbass gave the context and usually I think of this for the argument that $f' > 0$ implies strict monotonicity. –  Sam Feb 3 '13 at 12:12
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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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Nice hint: and nice observation! +1 –  amWhy Feb 3 '13 at 13:18
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