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**ppvanya** Hello guys, I have a problem with solving these inequalities. If anyone could help me I would really appreciate it.

1. e^x>1+ln(1+x), for all real numbers ( I understand how this is true for x>0, but somehow I can't seem to make it work for x<0, although it is obviously true, since f'(x)<0 for all x's except x=0.

We define $\displaystyle f(x)=e^x-\ln(x+1)-1$ . Since $\displaystyle \lim_{x\to -1}f(x)=\infty\,,\,\, f(0)=0\,\,\,and\,\,\,f'(x)=e^x-\frac{1}{x+1}=0\Longleftrightarrow x=0$ (why? Because $\displaystyle f'(x)$ is an increasing function) , we have that our function has a minimal point at x =0 and thus $\displaystyle \forall\,x\in (-1,\infty)\,,\,\,f(x) = e^x-\ln(x+1)-1\ge 0 = f(0)$

2. sinx<x- (x^3)/6+(x^5)/120, for x>=0 (this one doesn't seem quite right, since its graph kind of stays constant around x=2.5-3. Main question is how do I tell whether the derivative is positive or negative when it is a complicated function itself and it's hard to plug in numbers and check)

For this one you better check the power series (Taylor-MacClaurin polynomial) around zero of $\displaystyle \sin x$.

Tonio

Thank you in advance!