Show that $x^3y^3(x^3+y^3) \leq 2$

If $x,y$ are positive reals such that $x+y = 2$ show that $x^3y^3(x^3+y^3) \leq 2$.

I thought about working backwards on this one. We can try to prove that $x^3y^3(x^3+y^3) \leq 2$ is true by first realizing that $x^3+y^3 = (x+y)(x^2-xy+y^2)$ and substituting in for $x+y = 2$. We then get $x^3y^3(x^2-xy+y^2) \leq 1$. How do I prove this is true?

Solution 1:

Note that $x^2-xy+y^2=(x+y)^2-3xy=4-3xy.$
So, now you have to prove that $$(xy)^3(4-3xy) \leq 1,$$ which can be done as follows:

AM-GM inequality gives
$$\left(\frac{xy+xy+xy+(4-3xy)}{4}\right)^4 \geq (xy)^3(4-3xy)$$