<s>[INST] 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.
 [/INST] We will prove $$ab+bc+ca-2abc\le \frac{7}{27}$$ where $$\left(a;b;c\right)=\left(\frac{1}{3};\frac{1}{3};\frac{1}{3}\right)$$
Let $(a+b+c;ab+bc+ca;abc)\rightarrow (p;q;r) (p=1;q;r>0)$
So we can prove  $7p^3\ge 27\left(pq-2r\right)$
Or $7p^3+54r\ge 27pq$
By Schur'inequality: $$7p^3+54r\ge 7p^3+54\cdot \frac{p\left(4q-p^2\right)}{9}$$
Then it's enough to prove $7p^2+54\cdot \frac{\left(4q-p^2\right)}{9}-27q\ge0$
Or $$p^2-3q\ge 0\Leftrightarrow (a+b+c)^2\ge3(ab+bc+ca)$$ 
The last inequality is obvious.
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