<s>[INST] Cubic inequality If $x,y,z$ are reals from $[0,1]$, then prove that $$2(x^3+y^3+z^3)-x^2y-y^2z-z^2x \le 3$$ We can assume $x=sin\theta, y=sin\phi,z=sin\gamma$. Therefore inequality can be written as $$\begin{align}&2(sin^3\theta+sin^3\phi+sin^3\gamma)-sin^2\theta.sin\phi-sin^2\phi.sin\gamma-sin^2\gamma.sin\theta\le 3\\&=>4sin^3\theta+4sin^3\phi+4sin^3\gamma\le 6+2sin^2\theta.sin\phi+2sin^2\phi.sin\gamma+2sin^2\gamma.sin\theta\\&=>-sin3\theta-sin3\phi-sin3\gamma\le6+\sum_{cyc-\theta,\phi,\gamma}sin\theta(2sin^2\phi-3)\\&=>\sum_{cyc}sin\theta(2-cos2\phi)\le 6+\sum_{cyc}sin3\theta\end{align}$$ What do I do after this?? Is this approach correct?
 [/INST] Assume $x\geq y\geq z$. Then $x^2y\geq y^3, y^2 z\geq z^3,$ and $z^2x\geq z^3$. Consequently
$$
\begin{align*}
2(x^3+y^3+z^3)-x^2y-y^2z-z^2x&\leq 2(x^3+y^3+z^3)-y^3-z^3-z^3\\
&=2x^3+y^3\\
&\leq 3,
\end{align*}
$$
since $x,y\leq 1$.
EDIT: There was some consternation about why one is allowed to assume $x\geq y\geq z$. If it helps, here is the same argument phrased without making such an assumption.
Let $X$ denote the largest of $x,y,z$, let $Y$ denote the second largest, and let $Z$ denote the smallest.
$$
\begin{align*}
2(x^3+y^3+z^3)-x^2y-y^2z-z^2x&\leq 2(x^3+y^3+z^3)-Y^3-Z^3-Z^3\\
&=2(X^3+Y^3+Z^3)-Y^3-2Z^3\\
&=2X^3+Y^3\\
&\leq 3.
\end{align*}
$$
</s>