Prove that (e^x+e^-x)/2 > 1+ (x²)/2 for all x ≠0

Prove that $\dfrac{e^{x}+e^{-x}}{2}\gt 1+\dfrac{x^{2}}{2}$ for all $x\neq0$.

Step 1: The quickest route: the series

The even series of the hyperbolic cosine converges for every $x$ and has only positive terms: \[\cosh x = \frac{e^{x}+e^{-x}}{2} = 1+\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+\frac{x^{6}}{6!}+\cdots \tag{1}\]

Subtracting the first two terms leaves

\[ \cosh x-1-\frac{x^{2}}{2} = \frac{x^{4}}{24}+\frac{x^{6}}{720}+\cdots \gt 0 \qquad (x\neq0), \] since every term is positive when $x\neq0$.

Step 2: The same by differentiation

Put $f(x) = \cosh x-1-\dfrac{x^{2}}{2}$, so that

\[ f'(x) = \sinh x-x, \qquad f''(x) = \cosh x-1. \]

Step 3: Sign of the second derivative

By the arithmetic–geometric mean inequality, $\dfrac{e^{x}+e^{-x}}{2}\ge\sqrt{e^{x}e^{-x}} = 1$, with equality only at $x = 0$. Hence $f''(x)\gt 0$ for $x\neq0$ and $f''(0) = 0$.

Step 4: Sign of the first derivative, then of $f$

Since $f''\gt 0$ off the origin, $f'$ is strictly increasing on the whole line; and $f'(0) = \sinh0-0 = 0$. Therefore

\[ f'(x)\lt 0 \ \text{ for } x\lt 0, \qquad f'(x)\gt 0 \ \text{ for } x\gt 0, \] so $f$ decreases to $x = 0$ and increases after it: $x = 0$ is the unique global minimum, with $f(0) = 0$. Consequently $f(x)\gt 0$ for every $x\neq0$.

Step 5: Verification

\[x = 0.5:\ 1.1276260 \gt 1.125; \qquad x = 1:\ 1.5430806 \gt 1.5; \qquad x = 2:\ 3.7621957 \gt 3 \;\;\checkmark\]

\[\text{The gap is } \frac{x^{4}}{24}+\cdots:\ \text{at } x = 0.5,\ 0.0026042 \approx 1.1276-1.125 = 0.0026260 \;\;\checkmark\]

\[\text{Adding } e^{x}\gt 1+x \text{ and } e^{-x}\gt 1-x \text{ gives only } \cosh x\gt 1,\ \text{a weaker statement} \;\;\checkmark\]