Trying to prove that for x real, ln(1+x)<=x

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SUMMARY

The discussion centers on proving the inequality ln(1+x) ≤ x for real x. Participants suggest using the function f(x) = ln(1+x) - x and exploring its properties, particularly its concavity. One contributor emphasizes that the Taylor series expansion for e^x is sufficient for the proof, while another expresses interest in a concavity-based approach despite its complexity. The consensus is that while multiple methods exist, the Taylor series provides a straightforward solution.

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Homework Statement



Hi everyone,
I'm trying to proove that for x real, ln(1+x)<=x

Homework Equations





The Attempt at a Solution



I know that I can solve this problem by introducing f:=x->ln(1+x)-x and studying f, but I'd like to use x->ln(1+x)'s concavity so I tried to find lambda, x and y in :
ln(1+(lambda)*x+(1-lambda)*y)>=(lambda)*ln(1+x)+(1-lambda)*ln(1+y)...


Any help is welcome...
 
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You are trying too hard.

Do you know the taylor series expansion for e^x? That is all you need.
 


I find your solution quite interesting too (I had not thought about it) but I'd like to find a solution using the concavity of this function x:=->ln(1+x) (though I admit it may be too hard for that question).
 

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