# Proof of inequality $e^x + e^{-x} \leq 2e^{x^2}$

How would I prove that $$e^x + e^{-x} \leq 2e^{x^2}, \quad \text{for all real x}?$$

I narrowed it down to proving for $x \in (-1,1)$.

I observed that for $(0,1)$ and for $(-1,0)$ I may need to use different approximations. I tried using Taylor polynomials and Lagrange remainder but to no avail, would be interested in the solution using a Taylor series or Taylor polynomial if such exists.

• There is no need to worry about negative $x$, since the left side and the right are symmetric about the $y$-axis. Feb 11, 2013 at 18:13

$\displaystyle e^x+e^{-x} = \sum_n \frac{x^n}{n!} + \sum_n (-1)^n \frac{x^n}{n!} = 2\sum_{n \text{ even}} \frac{x^n}{n!} = 2 \sum_n \frac{(x^2)^n}{(2n)!} \leq 2 \sum_n \frac{(x^2)^n}{n!} = 2 e^{x^2}$.

• Oh, you've been faster. Feb 11, 2013 at 18:21
• A rare occurrence... Feb 11, 2013 at 18:22
• Can I actually do that for infinite sums (adding them together) isn't this reordering of summation? I guess you could, I was just warned of it, when can't I do something like that?
– NBP
Feb 11, 2013 at 18:29
• You can always add two convergent series together. There is no reordering above (although reordering would be valid in this case since the convergence is absolute). Feb 11, 2013 at 18:32
• @mahavir: Just saw this comment now. How do I respond? Mar 17, 2014 at 7:29

Recall $$e^x=\sum_{n\geq 0}\frac{x^n}{n!}\quad \forall x\in\mathbb{R}.$$ Now compute the series of $$e^x+e^{-x}=2\sum_{n\geq 0}\frac{x^{2n}}{(2n)!}\quad \forall x\in\mathbb{R}$$ and of $$2e^{x^2}=2\sum_{n\geq 0}\frac{x^{2n}}{n!}\quad \forall x\in\mathbb{R}.$$ Now compare.

We need only consider $x\ge 0$ as both sides are even functions.

The function $(0,\infty)\to\mathbb R, t\mapsto \frac{e^t-1}{t}$ is strictly increasing, takes the value $e-1<2$ at $t=1$ and approches $1$ as $t\to 0^+$. Therefore $$1+t\color{#008800}\le e^t\color{#ff0000}\le 1+2t\quad\text{for }0\le t\le 1.$$ (More generally: If $f$ is convex, then $f'(0)\le \frac{f(t)-f(0)}t\le f(1)-f(0)$ for $0< t\le 1$).

Hence for $0\le x\le 1$, we have $$e^x-2+e^{-x}=(e^{x/2}-1)^2\color{#ff0000}\le x^2\le 2x^2\color{#008800}\le 2(e^{x^2}-1)$$ as was to be shown. (And of course for $x\ge1$, we have $x^2\ge1$ and hence $2e^{x^2}\ge2e^x=e^x+e^x>e^x+e^{-x}$).

• Nice alternative! +1 Feb 12, 2013 at 0:08

I don't know how to type math site but here's a verbalized idea:

You can derive both sides, notice that both their first derivatives are positive over for all positive $x$ and that the functions are symmetric around $Y$ axis, therefore the global minimum of both are at $x=0$ where each side equals $2$. Comparing the derivatives you can also show that the right side grows faster for any $x \geq 0$ therefore $LHS \leq RHS$ for any $x \geq 0$.