<s>[INST] $e^{\frac{1}{x}} < 1 + \frac{1}{x-1} $ I want to prove that $e^{1/x} < 1 + \frac{1}{x-1}$ for $x > 1$.
The first thing I tried is differentiating $f(x) = e^{1/x} - 1 - \frac{1}{x-1}$: this gives 
$$ \frac{1}{x^2} \left( \left(1 + \frac{1}{x-1}\right)^2 - e^{\frac{1}{x}} \right) $$
If I could show that $\left(1 + \frac{1}{x-1} \right)^2 > e^\frac{1}{x} $ for $x>1$, then $f(x)$ would be increassing, and since $ \lim_{x \to \infty} f(x) = 0$ this would mean that $f(x) < 0$ for $x>1$. However, proving that inequality is very similar to the first one, and still involves bounding above $e^\frac{1}{x}$.
The other thing I tried is considering $f(x) = e^{x-1} - (x-1) - 1$ which is increasing for $x > 1$. Then $f(\frac{1}{x})$ is decreasing, so $e^{\frac{1}{x}-1} - \frac{1}{x}$ is decreasing;  However this also does not seem to help too much.
Any ideas?
 [/INST] $$e^{\frac{1}{x}}<\frac{x}{x-1}$$
$$e^{\frac{1}{x}}<\frac{1}{1-\frac{1}{x}}$$
$$e^{-\frac{1}{x}}<\frac{1}{1+\frac{1}{x}}$$
$$\frac{1}{e^{\frac{1}{x}}}<\frac{1}{1+\frac{1}{x}}$$
$$\frac{1}{1+\frac{1}{x}+\frac{1}{2x^2}+...}<\frac{1}{1+\frac{1}{x}}$$
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