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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.

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  • $\begingroup$ It is $$ab+bc+ca-2abc\le \frac{7}{27}$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 5 at 14:02
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    $\begingroup$ Dear Sonhard, why don't you post it as an answer? @Dr.SonnhardGraubner $\endgroup$ – Aqua Mar 5 at 14:12
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Fixing the value of $a$, we want to find the maximum of $$ S=ab+bc+ca-2abc=(1-2a)bc+a(1-a). $$ For $a>\frac 12$, we want to choose $b,c$ with $b+c=1-a$ that makes $bc$ as small as possible. The optimum can be achieved as one of $b$ and $c$ goes arbitrarily close to $0$, but this contradicts $b,c>0$ (But we need to check this case to make sure that $S$ indeed achieves it maximum in the interior). On the other hand, for $a\le \frac12$, we want to choose $b,c$ that makes $bc$ as large as possible. Given that $b+c=1-a$, the maximum value of $bc$ is given by AM-GM; $2\sqrt{bc}\le b+c=1-a\implies bc\le \frac{(1-a)^2}4$ with the equality attained when $b=c=\frac{1-a}2$. Inserting this, we have $$ S=\frac{(1-2a)(1-a)^2}4+a(1-a)=\frac{(1-a)(2a^2+a+1)}{4}. $$ By differentiating $S$ with respect to $a$, we have $$ S'=\frac{a(1-3a)}{2}. $$ Since $S'>0$ on $(0,\frac13)$ and $S'<0$ on $(\frac13,1)$, we know that $a=\frac 13$, $b=c=\frac{1-a}2=\frac 13$ is optimal. This gives $S\le \frac13-\frac2{27}=\frac7{27}$.

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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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Σ(ab) = Σ ( a²b+b²a ) + abc and by using AM-GM to the terms under the modified sigma we get that it is always greater than 6abc and using AM - GM on a,b,c we get abc<1/27 hence the maximal value must be 7/27

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  • $\begingroup$ i modified the expression by using 1 = a+b+c as given $\endgroup$ – Aditya Garg Mar 5 at 14:41
  • $\begingroup$ Best and shortest derivation in here ! $\endgroup$ – Andreas Mar 5 at 14:59
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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 $$ \frac{7}{27}>\frac 14 $$

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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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