Minimum and Maximum #2

Let $x+y=t,~x-y=s$ .

Then $t>0$ .

Since $x^3+x^2y-xy^2-y^3=(x-y)(x+y)^2=t^2s\ge12$ ,

$s\ge \dfrac{12}{t^2}$ .

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$\begin{aligned} 2x+y & =\frac{3}{2}t+\frac{1}{2}s \\ & \ge \frac{3}{2}t+\frac{6}{t^2} \\ & = \frac{3}{4}t+\frac{3}{4}t+\frac{6}{t^2} \\ & \ge 3 \sqrt[3]{\frac{3}{4}t\cdot\frac{3}{4}t\cdot\frac{6}{t^2}} \\ & = 3 \sqrt[3]{\frac{27}{8}} \\ & = \frac{9}{2} \\ \end{aligned}$

with equality achieved when $t=2,~s=3$ , in other words, $x=\dfrac{5}{2},~y=-\dfrac{1}{2}$ .

Therefore the minimum of $(2x+y)^2$ is $\dfrac{81}{4}=\boxed{20.25}$ .