How to prove this inequality involving exponential?

Question

For $0<\epsilon<1$, how to prove that $$1 + \epsilon \leq e^{\epsilon - \left(\epsilon^2 - \epsilon^3\right)/2}$$ Please help.

Answer 1

Knowing that $\ln$ is a function strictly increasing we can do

$$ \ln(1+\epsilon) \le \epsilon-\frac{\epsilon^2-\epsilon^3}{2} $$

Now expanding $\ln(1+\epsilon)$ in series around $\epsilon = 0, (0 < \epsilon < 1)$ and noting that this is an alternating convergent series we can proceed to demonstrate that.

Answer 2

For $x \in [0,1]$, consider $f(x) = e^{x - (x^2-x^3)/2}$ and $g(x) = 1+x$. Since $f(0) = g(0)$, to show what you want, we need $f'(x) \geq g'(x)$. If you take the first derivative, you will get $f'(x) = (1-x+\frac{3}{2}x^2) e^{x - (x^2-x^3)/2}$ which is not clearly greater than $g'(x) = 1$. However, we have once again $f'(0) = g'(0)$, so what you want next is $f''(x) \geq g''(x)$. And $f''(x) = [(1-x+\frac{3}{2}x^2)^2 + (3x - 1)]e^{x - (x^2-x^3)/2}$, which is greater than $ g''(x) = 0$. This is because the coefficient can be rewritten as $(3x^2 - 3x^3) + x + x^2 + \frac{9}{4}x^4 > 0$.

Answer 3

The derivative of $$ y=e^{x-\frac{x^2-x^3}{2}}$$ is equal to $$y'=e^{x-\frac{x^2-x^3}{2}}\left(\frac{3x^2-2x+2}{2}\right)$$

and then the tangent to the curve $ y=e^{x-\frac{x^2-x^3}{2}}$ in the point $(0,1)$ is precisely equal to $$y=1 + x$$ Since, as it is well known, the curve is convex we are done.

Answer 4

You just have to prove the corresponding inequality for the logs, i.e. $$\ln(1+\varepsilon)\le \varepsilon-\frac{\varepsilon^2} 2+\frac{\varepsilon^3}2. $$ Now Leibniz' theorem for alternating series shows that if $0<\varepsilon<1$, one has $$\ln(1+\varepsilon)\le\varepsilon-\frac{\varepsilon^2}2+\frac{\varepsilon^3}3,$$ so it all results from the inequality $\dfrac 13 <\dfrac12$, that you may happen to have known for some time…