Inequality doubt with taylor expansion

Question

Can I prove that $\forall x>0$ $$e^{x/(1+x)} < 1+x$$

Showing that $e^{x/(1+x)} = 1+x-\frac{x^2}{2}+o(x^2)$ and so $-\frac{x^2}{2}+o(x^2)<0$ for all $x>0$? How i can be sure that $o(x^2)$ doesn't interfere in the inequality also for large values of $x$?

Answer 1

Or using the Taylor expansion for $e^t$: $$e^{\frac{x}{x+1}} = 1+ \frac{x}{x+1} + \frac1{2!} \left(\frac{x}{x+1} \right)^2 + \cdots < 1+ \frac{x}{x+1} + \left(\frac{x}{x+1}\right)^2 + \cdots = \frac1{1-\frac{x}{x+1}} = 1+x$$

Answer 2

Taking logs of both sides, equivalently, you need to show that $f(x) \triangleq (1+x)\log(1+x)-x > 0$. The derivative is $f'(x) = \log(1+x) > 0,\,\forall x>0$, while $f(0) = 0$. The claim follows.