hello, we get $\frac{8(x+y)(x^3+y^3)}{(x^2+y^2)^2}\le 9$ and the equal sign holds if $x=-1,y=-2-\sqrt{3}$ Sonnhard.

How did you get $9$? What inequality did you use? What were your steps?

hello, we get $9-\frac{8(x+y)(x^3+y^3)}{(x^2+y^2)^2}=\frac{(x^2-4xy+y^2)^2}{(x^2+y^2)^2}\geq 0$ Sonnhard.

Oh, I see now. Thank you!

${\frac{8(x+y)(x^3+y^3)}{(x^2+y^2)^2}=\frac{4}{(x^2+y^2)^2}(x^2+2xy+y^2)(2x^2-2xy+2y^2)}$ $\le \frac{4}{(x^2+y^2)^2}\left(\frac{x^2+2xy+y^2+2x^2-2xy+2y^2}{2}\right)^2=9,$ and the equal sign holds if $x=-1,y=-2-\sqrt{3}$.

Let $x=r\cos t,y=r\sin t$ (this is a standart substitution when you see expressions involing $x^2+y^2$). Then $\frac{8(x+y)(x^3+y^3)}{(x^2+y^2)^2}=8(\cos t+\sin t)(\cos^3 t+\sin^3 t)=$ $=8(\cos t+\sin t)^2(1-\cos t\sin t)=8(1+2\cos t\sin t)(1-\cos t\sin t)$. Now let $u=\cos t \sin t$, so $-\frac{1}{2}\le u \le \frac{1}{2}$. It is easy to see that the maximum of $f(u)=8(1+2u)(1-u)$ is obtained when $u=\frac{1}{4}$ and is $8\frac{3}{2}\frac{3}{4}=9$.

Dr Sonnhard Graubner wrote: hello, we get $9-\frac{8(x+y)(x^3+y^3)}{(x^2+y^2)^2}=\frac{(x^2-4xy+y^2)^2}{(x^2+y^2)^2}\geq 0$ Sonnhard. But how did you just think "hey why don't I subtract that from 9?"

mathtastic wrote: But how did you just think "hey why don't I subtract that from 9?" how??? its a wolframalphmagic!...

sqing wrote: ${\frac{8(x+y)(x^3+y^3)}{(x^2+y^2)^2}=\frac{4}{(x^2+y^2)^2}(x^2+2xy+y^2)(2x^2-2xy+2y^2)}$ $\le \frac{4}{(x^2+y^2)^2}\left(\frac{x^2+2xy+y^2+2x^2-2xy+2y^2}{2}\right)^2=9,$ and the equal sign holds if $x=-1,y=-2-\sqrt{3}$. How did you managed to think of such a construction?

Use normalization?

$8\le \frac{8(x+y)(x^3+y^3)}{(x^2+y^2)^2}\le 9$.

Here is its general form