Prove that $x^3 + x^3 y^2 + x^2 y^3 + x^2 + y^2 + y^3 \geq 2xy(x+y+xy)$ Prove that $x^3 + x^3 y^2 + x^2 y^3 + x^2  + y^2 + y^3 \geq 
 2xy(x+y+xy)$ for $x,y \in \mathbb{R}^+$.
I started by multiplying everything out on the RHS to get the equivalent statement 
\begin{align*}
x^3 + x^3 y^2 + x^2 y^3 + x^2  + y^2 + y^3 \geq 
 2 x^2 y + 2 x y^2 + 2 x^2 y^2.
\end{align*}
I've tried numerous methods of attacking this problem, but I haven't had much luck yet. The closest I seem to have been able to get is to use the identity $(x-y)^2\geq 0$ to deduce that
\begin{align*}
x^3+x^3y^2\geq x^3+xy^2=x(x^2+y^2)\geq 2x^2y,
\end{align*}
and similarly, 
\begin{align*}
y^3x^2+y^3\geq x^2y+y^3=y(x^2+y^2)\geq 2xy^2.
\end{align*}
But now I am left with having to prove that $x^2+y^2\geq 2x^2y^2$, which is obviously false for $x,y \in \mathbb{R}^+$. I'm thinking I'm giving up too much trying to make the identity $(x-y)^2\geq 0$ work for this problem. 
I also tried writing everything as a polynomial in $x$ and showing that the resulting polynomial is positive for $x>0$, but this gets extremely messy. Does anyone have any ideas? Thanks in advance!

By AM-GM, we have $$x^3+y^3+x^2y^3+x^3y^2 \geq 4 \sqrt[4]{x^3y^3x^2y^3x^3y^2} = 4x^2y^2.$$ 
Similarly, we have  $$x^3y^2 + y^2 + x^2 + x^3 \geq 4 \sqrt[4]{x^3y^2y^2x^2x^3} = 4x^2y$$ and $$x^2y^3+ x^2 + y^2 + y^3 \geq 4 \sqrt[4]{x^2y^3x^2y^2y^3} = 4xy^2.$$ Adding these inequalities yields the desired result.
