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


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.


I also found an elegant proof in the editorial note in this old journal (from Jan 1938) that I stay here for completeness:

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 $f(x)>0$.


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