Proof that $\frac{e^x}{e^{\frac{x^2}{2}}} < 1+x$

Question

For $0<x<1$, prove that $\dfrac{e^x}{e^{\frac{x^2}{2}}} < 1+x$ holds.

I've tried rewriting this in numerous ways, for example by multiplying the RHS with the denominator of the LHS, however nothing so far has given me a result or something I can work with.

Answer 1

The derivative of $\ln (1+x)-x+\frac {x^{2}} 2$ is $\frac 1 {1+x} -1+x=\frac {x^{2}} {1+x}$ which is positive. Hence this functinn is increasing. Since its value when $x=0$ is $0$ we get $\ln (1+x)-x+\frac {x^{2}} 2 >0$. Take exponential to complete the proof.

Answer 2

A possible approach would be to prove that $$f(x)=e^{\frac{x^2}{2}} (x+1)-e^x$$ is positive if $x >0$.

Using Taylor expansion built at $x=0$, you would have $$f(x)=\frac{x^3}{3}+\frac{x^4}{12}+\frac{7 x^5}{60}+\frac{7 x^6}{360}+O\left(x^7\right)$$ and, if you continue the expansion, you should notice that all coefficients are positive. This means that the inequality holds for any $x >0$.