An algebra problem by Sukino

Since $x^2$ and $y^2$ are positive real, we can apply RMS-AM inequality and AM-GM inequality as follows.

RMS-AM inequality:

$\begin{aligned} \frac {x^2+y^2}2 & \le \sqrt {\frac {x^4+y^4}2} & \small \color{#3D99F6} \text{Equality occurs when } x = \pm 3, \ y = \mp 3 \implies x^2=y^2, \ x^2+xy+y^2 = 9 \\ & \le \sqrt{\frac {81+81}2} = 9 \\ \implies x^2+y^2 & \le 18 \quad \small \color{#3D99F6} (\text{maximum}) \end{aligned}$

AM-GM inequality:

$\begin{aligned} x^2+y^2 & \ge 2\sqrt{x^2y^2} = 2xy & \small \color{#3D99F6} \text{Equality occurs when } x = y = \sqrt 3 \implies x^2+xy+y^2 = 9 \\ & \ge 6 \quad \small \color{#3D99F6} (\text{minimum}) \end{aligned}$

$\implies \text{maximum + minimum} = 18+6=\boxed{24}$

$x^2+xy+y^2=9 \\ (x-y)^2+2xy+xy=9 \\ (x-y)^2+3xy=9 \\ (x-y)^2=9-3xy \geq 0 \\ xy \leq 3$

Equality holds when $x=y$

So , minimm value is $x^2+y^2=9-xy=9-3=6$

$x^2+xy+y^2=9 \\ (x+y)^2-2xy+xy=9 \\ (x+y)^2-xy=9 \\ (x+y)^2=9+xy \geq 0 \\ xy \geq -9$

Equality holds when $x=-y$

So,maximum value of $x^2+y^2=9-xy=9+9=18$

The given curve is an ellipse with centre at the origin, so any point $(r\cos\theta, r\sin\theta)$ can be taken on it

which means $x^{2} + y^{2} = (r\cos\theta)^{2} + (r\sin\theta)^{2} = r^{2}$ . after satisfying this point on the curve it yields

$r^{2} = \frac{18}{2 - \sin2\theta}$

which gives max. $r^{2} = 18$ and min. $r^{2} = 6$ $\Rightarrow$ $18 + 6 = \boxed{24}$ .