Inequality $x^4+y^4+(x^2+1)(y^2+1)\ge x^3(1+y) +y^3(1+x)+x+y$ for $x,y \in\mathbb{R}$ Prove for $x,y \in\mathbb{R}$ that such inequality exists ;
$x^4+y^4+(x^2+1)(y^2+1)\ge x^3(1+y) +y^3(1+x)+x+y$
And here is what I realised ;
because $(x^2+1)(y^2+1) >=1$
and $x^4+y^4 \ge 0$
$1\ge x^3(1+y) +y^3(1+x)+x+y$
I'd prefer a hint

Hint: First expand the inequality. Then expand $(x-y)^4$ and use it to get rid of the $x^3y+xy^3$ terms. Beat the rest with repeated usages of AM-GM in the form $a^2+b^2\ge 2ab$ with various choices of $a,b$.


A short solution: It's equivalent to
$$\frac14(x-y)^4+\frac14(x^2-y^2)^2+\frac12(x^2+1)(x-1)^2+\frac12(y^2+1)(y-1)^2,$$
which is obviously greater or equal to zero.
More detail:
After expanding the terms and multiplying by 2, your inequality becomes
$$2x^4+2y^4+2x^2y^2+2x^2+2y^2+2\ge 2x^3y+2xy^3+2x^3+2y^3+2x+2y$$
Note that $(x-y)^4+(x^2-y^2)^2\ge 0\implies x^4+y^4+2x^2y^2\ge 2x^3y+2xy^3$, so it suffices to show
$$x^4+y^4+2x^2+2y^2+2\ge 2x^3+2y^3+2x+2y$$
We have $(x-1)^2+(y-1)^2\ge 0\implies x^2+y^2+2\ge 2x+2y$, so it reduces to
$$x^4+y^4+x^2+y^2\ge 2x^3+2y^3\iff x^2(x-1)^2+y^2(y-1)^2\ge 0$$
