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

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).
