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?