Prove that $abc(a^2+b^2+c^2-2abc-2a-2b-2c)+ab(a+b-2)+bc(b+c-2)+ca(c+a-2)+a+b+c \le 2$ Prove that if the real numbers $a,b,c$ lie in the interval $[0,1]$,
then: 
\begin{align*}
abc(a^2+b^2+c^2-2abc-2a-2b-2c)+ab(a+b-2)+bc(b+c-2)+ca(c+a-2)+a+b+c \le 2.
\end{align*}
I've tried this:
\begin{align*}
abc(a^2+b^2+c^2-2abc-2a-2b-2c)+ab(a+b-2)+bc(b+c-2)+
\end{align*}
\begin{align*}
+ ca(c+a-2)+(a+b+c-2) \le abc(a^2+b^2+c^2-2abc-2a-2b-2c)+
\end{align*}
\begin{align*}
+ab(2-2)+bc(2-2)+ca(2-2)+(3-2)=
\end{align*}
\begin{align*}
=abc(a^2+b^2+c^2-2abc-2a-2b-2c)+1
\end{align*}
but that $1$ in the last line ruins it even though that first term is clearly negative. Can anyone help me?

$$\sum_{cyc}ab(a+b-2)\leq0$$ and
$$abc\sum_{cyc}(a^2-a)\leq0.$$
Thus, it's enough to prove that
$$a+b+c-abc(a+b+c)\leq2.$$
Now, let $a=\frac{1}{1+x},$ $b=\frac{1}{1+y}$ and $c=\frac{1}{1+z},$ where $x$, $y$ and $z$ are non-negatives.
Thus, we need to prove that
$$2(1+xyz)^2+\sum_{cyc}(3x^2y^2z+x^2y^2+4x^2yz+x^2y+x^2z+xy+x)\geq0,$$ which is obvious.
