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Find the minimum value of $(\frac{x^n}{n} + \frac{1}{x})$ for $n \ge 4$.

One possible approach could be by first writing $ \left(\frac{x^n}{n} + \frac{1} {x}\right) = \left( \frac{x^n}{n} + \frac{1}{nx} + \frac{1}{nx}+ \frac{1}{nx} + \text{upto n terms}\right) $ then using the property $\mathrm{AM} \ge \mathrm{GM}$, we would get $\left(\frac{x^n}{n} + \frac{1}{nx} + \frac{1}{nx}+ \frac{1}{nx}+ \text{upto }n \text{ terms}\right) = \left( \frac{x^n}{n} + \frac{1}{x} \right) \ge \frac{n+1}{n}$ But this does not hold good for negative $x$; I am inquisitive to know how to approach for that case?

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    The AM-GM approach is indeed nice. But the calculus deals just as easily, for example, with the minimum value of $\frac{x^{\sqrt{2}}}{\sqrt{2}}+\frac{1}{x}$ for positive $x$.2011-08-02

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For $x \lt 0$ there is no minimum. As $x$ gets very close to $0$, the $\frac{1}{x}$ term gets very large and negative, faster than the $\frac{x^n}{n}$ can get positive (assuming $n$ is even-if $n$ is odd this term is negative, too).

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    It's a joke! You're supposed to put spaces after punctuat$i$on. No biggie.2011-08-01