Slick proof of exponential inequality

Question

Today I saw that using taylor series, one can show that $e^x+e^{-x}\leq 2e^{x^2/2}$. Is there a slick proof using some sort of Jensen-type inequality or integral bound?

Answer 1

As @flawr points out, the L.H.S. is $\displaystyle \cosh x = \frac{e^x+e^{-x}}{2}$ has an infinite product representation:

$$\cosh x = \prod\limits_{n=1}^{\infty} \left(1+\frac{4x^2}{\pi^2(2n-1)^2}\right) \le \exp \sum\limits_{n=1}^{\infty}\frac{4x^2}{\pi^2(2n-1)^2} = e^{x^2/2}$$

Answer 2

$y = 2 e^{x^2/2}$ satisfies the d.e. $y' = x y$ with $y(0) = 2$, while $z = e^{x} + e^{-x} = 2 \cosh(x)$ satisfies $z' = \tanh(x) z$ with $z(0)=2$. Since $x \ge \tanh(x) $ for $x \ge 0$, Gronwall's inequality does the rest.

Answer 3

First, since both sides are even functions, it is sufficient to prove that $e^x+e^{-x} \leq 2e^{x^2/2}$ for all $x \geq 0$. Then we can change the problem to the equivalent problem of proving that for $x \geq 0$ $$ u(x) \equiv \frac{1}{e^x+e^{-x}} \geq \frac{1}{2}e^{-x^2/2} \equiv v(x)$$ We start by observing that $u(0) = v(0).$

Now in the immortal words of William Feller, let's pass to the logarithm: We will aim at showing that for $\forall x > 0$, $$\ln u(x) > \ln v(x)$$. To do this, we will start with $\ln u(0) > \ln v(0)$ and note that $$ \frac{d}{dx} \ln u(x) = \frac{1}{u} d\frac{du(x)}{dx} = -\frac{e^x-e^{-x}}{e^x+e^{-x}} = -\frac{e^x+e^{-x}}{e^x+e^{-x}} +\frac{2e^{-x}}{e^x+e^{-x}} $$ $$ \left. \frac{d}{dx} \ln u(x) \right|_{x=0} =0 $$ $$ \frac{d}{dx} \ln v(x)=\frac{1}{v} d\frac{dv(x)}{dx} = -x$$ $$ \left. \frac{d}{dx} \ln v(x) \right|_{x=0} =0 $$ $$ \frac{d^2(\ln u(x))}{dx^2} = -\, \frac{4}{(e^x+e^{-x})^2}> -1 $$ $$ \frac{d^2(\ln v(x))}{dx^2} = -1$$ So the functions $\ln u(x)$ and $\ln v(x)$, along with their derivatives, match at $x=0$, and for all $x>0$ the second derivative of $\ln u(x)$ is greater than the second derivative of $\ln v(x)$. Thus $$\ln u(x) \geq \ln v(x) \Rightarrow u(x) \geq v(x) \Rightarrow \frac{1}u(x) \leq \frac{1}{v(x)} \Rightarrow e^x + e^{-x} \leq e^{-x^2/2} $$ which is what we set out to prove. By using the second derivative we effectively used the concavity of the function $\ln u - \ln v$.

Answer 4

Note that by the Mean Value Theorem, we have $$ \frac{\tanh(x)}{x}=\mathrm{sech}^2(\xi)\tag{1} $$ for some $\xi$ between $0$ and $x$. Therefore, for $x\ne0$, $$ 0\lt\frac{\tanh(x)}{x}\lt1\tag{2} $$ Thus, for $x\ne0$, $$ \begin{align} x\frac{\mathrm{d}}{\mathrm{d}x}\left(e^{x-x^2/2}+e^{-x-x^2/2}\right) &=(x-x^2)e^{x-x^2/2}-(x+x^2)e^{-x-x^2/2}\\ &=e^{-x^2/2}\left(2x\sinh(x)-2x^2\cosh(x)\right)\\ &=2x^2e^{-x^2/2}\cosh(x)\left(\frac{\tanh(x)}{x}-1\right)\\ &\lt0\tag{3} \end{align} $$ If $xf'(x)\lt0$ for $x\ne0$, then $f$ has a global maximum at $x=0$. Thus, $$ e^{x-x^2/2}+e^{-x-x^2/2}\le2\tag{4} $$ which is the same as what is asked.