# Proof of an exponential inequality

This seems to be obvious but I am having a hard time proving it. Any insight greatly appreciated.

Statement:

Prove for every $$b,x \in \mathbb{R}$$ such that $$b\geq 1$$, $$|x|\leq b$$, it holds that $$(1+\frac{x}{b})^b \geq e^{x}(1-\frac{x^2}{b}).$$

My attempt:

Some cases are trivial: if $$x = 0$$ it holds with equality, or if $$x^2\geq b$$ the RHS is negative while the LHS is always positive. For $$b = 1$$, I was able to show it by separating $$x>0$$ and $$x<0$$ cases. When $$b = 1$$ and $$x>0$$, the function $$h_b(x) = (1+\frac{x}{b})^b-e^{x}(1-\frac{x^2}{b})$$ can be easily shown to be non-decreasing by taking the derivatives and employing $$e^{-x} \geq 1-x$$ establishes the result. For $$b = 1$$ and $$x<0$$ the function is neither decreasing nor increasing so we can't take the derivative. Instead, I directly worked with $$h_b(x)$$ and used $$e^x\leq 1+x+ \frac{x^2}{2}$$ for all $$x<0$$ to establish the results.

When $$b>1$$, I observed using plots that for $$x>0$$ the function is increasing and for when $$b\geq2$$ the function is decreasing for $$x>0$$. So theoretically we can prove the inequality by taking the derivatives. But for $$b >1$$ the derivatives are very involved. I also had a failed attempt using Taylor's inequality.

The left hand side is always non-negative, the right hand side is non-positive for $$|x|\ge \sqrt b$$. Hence we need only consider the case $$|x|<\sqrt b$$.
Apply log to the claim and rearrange: $$\tag1 b\ln\left(1+\frac xb\right)- x-\ln\left(1-\frac {x^2}b\right)\stackrel?\ge 0$$
Check the derivative: $$\frac1{1+\frac xb}-1+\frac{2x}b\frac1{1-\frac{x^2}b} =-\frac{x}{b+x}+\frac{2x}{b-x^2} =\frac{2x^2+bx+x^3}{(b+x)(b-x^2)}=\frac {x((x+1)^2+b-1)}{(b+x)(b-x^2)}.$$ As $$b\ge1$$ and $$|x|< \sqrt b\le b$$, the denominator is positive and the second factor in the numerator is positive (clearly so if $$b>1$$, while for $$b=1$$, we note that $$x=-1$$ is excluded). Hence the sign of the derivative is the same as the sign of $$x$$, i.e., the left hand side of $$(1)$$ is strictly decreasing for negative $$x$$ and strictly increasing for positive $$x$$. As $$x=0$$ makes the left hand side zero, the inequality questioned in $$(1)$$ holds for all $$x\in(-\sqrt b\sqrt b)$$, as desired.
Consider $$f(x) = e^{-x}(1+\frac{x}{b})^b-(1-\frac{x^2}{b})$$. At $$x = 0$$, both $$f(x)$$ and $$f^\prime(x)$$ are zero. If $$f^\prime(x) = 0$$ for any other $$x$$ in the interval, for such $$x$$ we have $$e^{-x}(1+\frac{x}{b})^b = 2+\frac{2x}{b}.$$ Therefore, for such $$x$$ $$f(x) = \frac{(x+1)^2}{b}+1-\frac{1}{b}>0.$$ Furthermore, since $$f(b)>0$$ for all $$b$$ while $$f(-b)>0$$ for $$b>1$$ and $$f(-b)=0$$ for $$b=1$$, all other points we must have $$f(x)>0$$.