maximum value of $\sum ab-2abc$ If $a+b+c=1$ and $a,b,c\in(0,1)$, then what is the maximum value of $(ab+bc+ca-2abc)$?
What I've tried:
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 4(ab+bc+ca)$
$(a-b)^2=a^2+b^2-2ab\geq 0$
$a^2+b^2\geq 2ab,b^2+c^2\geq 2bc,c^2+a^2\geq 2ca$
$ab+bc+ca\leq\frac14$
How do I solve it help me please.

Let $a, b, c\in [0, 1]$ satisfying all the hypothesis and $a=\max\{a, b, c\}>b$, it's clear that this expression must have a maximum. Set
$$
a'=\frac{a+b}2<a\\
b'=\frac{a+b}2>b
$$
then $a'+b'+c=1$ and
$$
a'b'+a'c+b'c=\frac{(a+b)^2}4+ab+bc=ab+bc+ac+\frac{(a-b)^2}{4}\\
2a'b'c=2c\frac{(a+b)^2}{4}=2abc+2c\frac{(a-b)^2}4\\
a'b'+a'c+b'c-2a'b'c=ab+bc+ac-2abc+(1-2c)\frac{(a-b)^2}{4}
$$
If $c\geq \frac{1}{2}$ then $a=c=\frac{1}{2}$ and $b=0$ so $ac=\frac 14$. Otherwise
$$
ab+bc+ac-2abc<a'b'+b'c+a'c-2a'b'c
$$
and the maximum should be $a=b=c=\frac 13$. The proof is concluded observing that
$$
\frac{7}{27}>\frac 14
$$
