AM-GM + Complex!

Rewrite $y = \large \frac{x^4 + x^2 +4}{x}$ as $\large x^3 + x + \dfrac 4 x$

$\large = x^3 + x +\dfrac 1 x +\dfrac 1 x +\dfrac 1 x +\dfrac 1 x$

Since $x \in (0,\infty)$ , using AM-GM Inequality

$\large \frac{x^3 + x +\dfrac 1 x +\dfrac 1 x +\dfrac 1 x +\dfrac 1 x }{6}$

$\large \geq (x^3 . x .\dfrac 1 x .\dfrac 1 x .\dfrac 1 x .\dfrac 1 x)^{\dfrac 1 6}$

$y_{min} = 6$

Using results of Triangular Inequality ,

$|z_1 + z_2| \geq ||z_1| - |z_2||$

& $|z_1 + z_2| \leq |z_1| + |z_2|$

we get $\alpha = 4$ and $\beta = 6$

Or one can interpret it geometrically also, given equation is a circle with center at $(4,-3)$ and radius $1$ .

Hence Maximum Distance of circle from origin = Radius + Distance of center from origin .

Minimum distance of circle from origin = - Radius + Distance of center from origin.

The complex equation is that of a disc of radius 1 and centre (4,-3). Since |z| represents distance from origin, The max. distance is distance of centre+radius and min. distance is distance of centre-radius. So, max. =6 ; min.=4; The value of k can be found by differentiating the equation finding x greater than zero and it comes out to be 6. So, k=max. value i.e., 6