Prove that if $x$ and $y$ are both not $0$ Prove that if $x$ and $y$ are both  not $0,$ then 
$$x^4+x^3y+x^2y^2+xy^3+y^4>0$$
I know this seems fairly easy but I'm fairly new to calculus and need some help proving that this is true. Appreciate the help!

Another way:
Note that $0\leq x^2(x+y)^2=x^4+2x^3y+x^2y^2$, and similarly $y^4+2xy^3+x^2y^2\geq 0$. Adding these and dividing by $2$, we get $\frac12x^4+x^3y+x^2y^2+xy^3+\frac12y^4\geq 0$. Then add $\frac{x^4+y^4}{2}$, which is positive since $x$ and $y$ are not both $0$.
