Write $x=1-y$, and then we have to maximize a function of one variable $y$.

khan.academy wrote: Problem Section #2 b) Find the maximal value of $(x^3+1)(y^3+1)$, where $x,y \in \mathbb{R}$, $x+y=1$. Set K equals to maximum value Then when we bash out: $$k\ge x^3y^3+y^3+x^3+1$$Then we for AM-GM we know: $$x+y\ge2\sqrt{xy}$$$$\frac{1}{4}\ge xy$$Then we also know: $$(x+y)^3=x^3+y^3+3xy(x+y)$$Or $$1-3xy=x^3+y^3$$Hence we have: $$k\ge 1+1-3xy+x^3y^3$$Or $$k\ge 2+xy(x^2y^2-3)$$Or $$k\ge 2+\frac{1}{4}(\frac{1}{16}-3)$$So we’re done...!

That is easy. But I think we have larger value than that.

khan.academy wrote: Yeah you got correct, but I have more simpler way than this which follows Am-Gm. Lol it's like 6 months, I didn't do inequalities so...

khan.academy wrote: Problem Section #2 b) Find the maximal value of $(x^3+1)(y^3+1)$, where $x,y \in \mathbb{R}$, $x+y=1$. $$(x^3+1)(y^3+1)\leq 4$$

Please elaborate!

khan.academy wrote: Problem Section #2 b) Find the maximal value of $(x^3+1)(y^3+1)$, where $x,y \in \mathbb{R}$, $x+y=1$. $$(x^3+1)(y^3+1)=(1-xy)^2(xy+2)\leq \frac{1}{2}\left(\frac{2xy+4+1-xy+1-xy}{3}\right)^3=4$$When $xy=-1$... (Zhangyanzong)

Ooops sorry!

khan.academy wrote: I think for $x,y\in \mathbb{R}$, $x+y=1,xy=-1$ is not possible. $$\{x, y\}=\{ \frac{1-\sqrt 5}{2},\frac{1+\sqrt5}{2} \}$$

@khan.academy You could easily get that using quadratic function.

sqing wrote: khan.academy wrote: Problem Section #2 b) Find the maximal value of $(x^3+1)(y^3+1)$, where $x,y \in \mathbb{R}$, $x+y=1$. $$(x^3+1)(y^3+1)=(1-xy)^2(xy+2)\leq \frac{1}{2}\left(\frac{2xy+4+1-xy+1-xy}{3}\right)^3=4$$When $xy=-1$... (Zhangyanzong) sqing wrote: khan.academy wrote: Problem Section #2 b) Find the maximal value of $(x^3+1)(y^3+1)$, where $x,y \in \mathbb{R}$, $x+y=1$. $$(x^3+1)(y^3+1)=(1-xy)^2(xy+2)\leq \frac{1}{2}\left(\frac{2xy+4+1-xy+1-xy}{3}\right)^3=4$$When $xy=-1$... (Zhangyanzong) Could you please elaborate which inequality are you using to get second relation (of inequality)?

It's just AM-GM, see carefully

khan.academy wrote: Problem Section #2 b) Find the maximal value of $(x^3+1)(y^3+1)$, where $x,y \in \mathbb{R}$, $x+y=1$. Is calculus allowed? Then we can just differentiate