Prove that $abc(a^2+b^2+c^2-2abc-2a-2b-2c)+ab(a+b-2)+bc(b+c-2)+ca(c+a-2)+a+b+c \le 2$

Question

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?

Answer 1

$$\sum_{cyc}ab(a+b-2)\leq0$$ 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.