<s>[INST] How do we prove without calculus that $\forall \ x,y \ge 0$, we have $\ 1+x^3+y^3\ge x+x^2y+y^2$ I've been trying to prove an inequality I was given by a friend, but so far my only progress has been calculus bashing:
$$LHS \ge RHS\iff1+x^3+y^3 - x-x^2y-y^2 \ge0$$
Letting $f(x,y) = 1+x^3+y^3 - x-x^2y-y^2$, we want $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$ for the minimum.
Hence $$3x^2-1-2xy=0 \ \ \ \ \ \ \ \ ... (1)$$ and $$3y^2-x^2-2y=0  \ \ \ \ \ \ \ \ ... (2)$$
which gives one solution$,(x,y)=(1,1),$ in the first quadrant:

The trouble is that proving that $(x,y)=(1,1)$ is the only positive solution to the system of equations is quite cumbersome.
$$\ \ $$
Is there any way we can prove this without calculus, maybe using AM-GM or Cauchy-Schwarz inequalities?
I used AM-GM to obtain $$1+x^3+y^3 \ge 3\sqrt[3]{1^3x^3y^3}=3xy,$$
but I'm not sure how this helps me proceed. Many thanks in advance.
 [/INST] Note that
$$
\begin{align}
\frac{1}{3}(1 + 1+x^3) &\ge x \\
\frac{1}{3}(x^3 + x^3 + y^3) &\ge x^2y \\
\frac{1}{3}(1 + y^3 + y^3) &\ge y^2 
\end{align}
$$
</s>