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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.

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

4 Answers 4

up vote 20 down vote accepted

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

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Oh, you've been faster. –  1015 Feb 11 '13 at 18:21
    
A rare occurrence... –  copper.hat Feb 11 '13 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 '13 at 18:29
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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). –  copper.hat Feb 11 '13 at 18:32
    
@mahavir: Just saw this comment now. How do I respond? –  copper.hat Mar 17 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.

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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 $2e^{x^2}\ge2e^x=e^x+e^x>e^x+e^{-x}$).

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Nice alternative! +1 –  1015 Feb 12 '13 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 $x \geq 0$.

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