# Lagrange theorem help me?

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 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 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 $0 ## 3 Answers Find the minima of the functions$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$x > \ln (1+x)$. - (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... -  Nice answer, Don. + – Babak S. Feb 3 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 at 11:54 It's simpler, imo, as it happens that$\,f'>0\,\,\,\forall x>0\,$– DonAntonio Feb 3 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 at 12:12 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 at 13:18