Prove $x^4+y^4+x^2+y^2+x^3y+y^3x\geq 0$ I tried 2 ways, first, take $x$ and $y$ $\geq0$ then obviously true. Take $x$ and $y$ both $\leq0$ same thing.
Now $y<0<x$ and wlog $|y|\leq x,$ then for big $x$ small $|y|$ we have $-x^4\leq x^3y$ and $-x^2\leq y^3x$ but if both $|y|$ and $x$ are big I got stuck.
So I decided to do it regular way and I looked for the maximum/maximas finding partial derivatives and setting it to zero and I got $$
\begin{cases}
  0=-4x^3+-2x+3x^2y-y^3 \\ 0=-4y^3-2y-3y^2x-x^3\\
\end{cases}
$$
easy to see (0,0) is a solution however I do not know how to prove that it is the only one
 A: EDIT
As my first answer was mistaken, a correction is coming now.
$$\begin{aligned}x^4+y^4+x^2+y^2+x^3y+xy^3&=\underbrace{x^4+y^4{\color{red}{-2x^2y^2}}}+x^2+y^2\underbrace{{\color{red}{+2x^2y^2}}+xy(x^2+y^2)}\\
&= (x+y)^2(x-y)^2  +x^2+y^2 \quad +xy(x+y)^2 \\
&=(x+y)^2\left((x-y)^2+xy\right)+x^2+y^2\\&=(x+y)^2(x^2-xy+y^2)+x^2+y^2\\
&=(x+y)^2\left((x-{y\over 2})^2+\frac{3y^2}{4}\right)+x^2+y^2\end{aligned}$$
which is non-negative.
A: Multiply by $4$ and use $(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$:
$$4(x^4+y^4+x^2+y^2+x^3y+y^3x)\geq 0 \iff \\
(x+y)^4+4x^2+4y^2+\underbrace{3(x^4+y^4)-6x^2y^2}_{\ge 0 \ by \ AM-GM}\ge 0.$$
A: $ab \leq \frac 1 2(a^{2}+b^{2})$ for any $a, b \in \mathbb  R$. This gives $xy \geq -\frac 1 2 ((x^{2}+y^{2}))$ and $x^{2}y^{2} \leq \frac 1 2 (x^{4}+y^{4})$. Use the first inequality for in the part $x^{3}y+xy^{3}=xy(x^{2}+y^{2})$ and, after simplification, finish the proof using the second inequality.
A: divide by $x^2y^2$if $x\neq0$ and $y\neq0$
$$\frac{x^2}{y^2}+\frac{y^2}{x^2}+\frac{1}{y^2}+\frac{1}{x^2}+\frac{x}{y}+\frac{y}{x}\geq 0$$
$$(\frac{x}{y}+0.5)^2+(\frac{y}{x}+0.5)^2+\frac{1}{y^2}+\frac{1}{x^2}-0.5\geq 0$$
now if $x\geq y$ then $(\frac{x}{y}+0.5)^2>1$
or if $y\geq x$ then $(\frac{y}{x}+0.5)^2>1$
