# Inequality proof for $1+x^3\geq x+x^2$

I have an inequality to prove and I can't get a hold of it... I hope someone can help with it or point me in the right direction. I tried it based on previous one, but without success... The prev. ones also seemed easy after I grasped them, so I'm sure that it's just a bit hard for me to get the first thought on the right path...

$$x\in\mathbb{R}, x\geq 0$$ $$1+x^3\geq x+x^2$$

\begin{align} 1+x^3\geq x+x^2 &\iff x^3-x \geq x^2 - 1\\ \\ & \iff x(x^2-1) \geq x^2 - 1 \quad\forall x\geq 0\end{align}
$1+x^3\geq x(1+x)$ for all $x$ with $x>0$, then we have $1+x^3-(1+x)x\geq 0$
$(1+x)(1+x^2-x)-x(1+x)\geq 0$ this is equivalent to
$(1+x)(1+x^2-2x)\geq 0$ or $(x-1)^2(x+1)\geq 0$ which is true. Sonnhard.
Hint: Both sides are multiples of $1+x$.