Prove : $0\leq ab+bc+ca-2abc \leq \frac{7}{27}$ $a;b;c\geq 0$ such that : $a^2+b^2+c^2=1$. 

Prove : $0\leq  ab+bc+ca-2abc \leq \frac{7}{27}$

Thanks :)
P/s : I have no ideas about this problem :(

This is the incorrect problem. The correct problem is 
Given $a,b,c \ge 0, a+b+c = 1,$ prove $0 \le ab+bc+ca-2abc \le \frac{7}{27}$.
For the LHS, $ab \ge abc$ and so on so we are done.
For the RHS, $\sqrt{xy} \le x+y = 1-z $ then $z(x+y)+xy(1-2z) \le z(1-z)+\frac{(1-2z)(1-z)^2}4 \le \frac{7}{27}$ by calculus methods. 
