<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] For $a=b=c=\frac{1}{3}$ we get the value $\frac{7}{27}.$
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$(a+b+c)(ab+ac+bc)-2abc\leq\frac{7(a+b+c)^3}{27}$$ or
$$\sum_{cyc}(7a^3-6a^2b-6a^2c+5abc)\geq0$$ or
$$7\sum_{cyc}(a^3-a^2b-a^2c+abc)+\sum_{cyc}(a^2b+a^2c-2abc)\geq0$$ or
$$7\sum_{cyc}(a^3-a^2b-a^2c+abc)+\sum_{cyc}c(a-b)^2\geq0,$$
which is true by Schur.
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