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.

• It is $$ab+bc+ca-2abc\le \frac{7}{27}$$ – Dr. Sonnhard Graubner Mar 5 at 14:02
• Dear Sonhard, why don't you post it as an answer? @Dr.SonnhardGraubner – Aqua Mar 5 at 14:12

5 Answers

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 $$S\le \frac13-\frac2{27}=\frac7{27}$$.

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.

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 $$a'=\frac{a+b}2b$$ 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 $$ab+bc+ac-2abc and the maximum should be $$a=b=c=\frac 13$$. The proof is concluded observing that $$\frac{7}{27}>\frac 14$$

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.

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

• i modified the expression by using 1 = a+b+c as given – Aditya Garg Mar 5 at 14:41
• Best and shortest derivation in here ! – Andreas Mar 5 at 14:59