<s>[INST] Inequality involving an exponent I wish to prove the following inequality
$$x^{\frac{3}{x-1}} > 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}, \quad x > 1.$$
Graphically the above inequality appears to be true since if one plots 
$$g(x) = x^{\frac{3}{x-1}} - \left (1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} \right )$$
it appears as though $g(x) > 0$ for all $x > 1$. I have however been unable to prove analytically this is true. 
I know 
$$1 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} < 4 \quad \mbox{for all} \,\, x> 1$$
and
$$1 < x^{\frac{3}{x - 1}} < \mathrm{e}^3 \quad \mbox{for all} \,\, x > 1,$$
but neither of these bounds seem to help me very much.
Any pointers in the right direction would be greatly appreciated. 
 [/INST] Hint :
Consider the Taylor expansion at infinity of your exponent :
$$x^{\frac{3}{x-1}} = x ^{3\ \big(\frac{1}{x} +\frac{1}{x^2} + \frac{1}{x^3} +...\big)}$$
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