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Given that $a,b,c$ are the lengths of the three sides of a triangle, and $ab+bc+ac=1$, the question is to prove $$(a+1)(b+1)(c+1)\leq4\,.$$

Any idea or hint would be appreciated.

This is Problem 6 of Round 1 of the BMO (British Mathematical Olympiad) 2010/2011, as can be seen here.

Remark. This question has been self-answered. Nevertheless, any new approach is always welcome!

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  • $\begingroup$ I have found another way how to solve your problem. If you want to see my solution, show please your attempts. $\endgroup$ Aug 22, 2020 at 5:55
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    $\begingroup$ @MichaelRozenberg The attempt by the OP is in one of the answers. See the remark I made in the question. $\endgroup$ Aug 22, 2020 at 6:06
  • $\begingroup$ @Batominovski Thank you! Now I see it. I just saw that someone wants to delete this topic... $\endgroup$ Aug 22, 2020 at 6:09
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    $\begingroup$ @MichaelRozenberg Then we shall undelete and reopen it if need be. This question has all the requirements. Context (source) and attempt. $\endgroup$ Aug 22, 2020 at 6:11

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Hint: Expanding the LHS gives us $(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc+1.$

Now, $(1-a)(1-b)(1-c)=1+ab+bc+ca-a-b-c-abc$.

Adding both the identities, we get $$\prod_{cyc}(1+a)+\prod_{cyc}(1-a)=4$$

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    $\begingroup$ That's very clever @SarGe. Now I know how to complete the proof :). We need to be able to show $(1-a)(1-b)(1-c)$ is positive right? $\endgroup$ Aug 20, 2020 at 11:53
  • $\begingroup$ And of course showing that $(1-a)(1-b(1-c)$ is positive is where we use the information that $a$ $b$ and $c$ are the sides of the triangle which has not been used yet in the steps you have shown... $\endgroup$ Aug 20, 2020 at 11:57
  • $\begingroup$ @SimonTerrington, you got it right. $\endgroup$
    – SarGe
    Aug 20, 2020 at 12:04
  • $\begingroup$ What an ingenious approach! By the way, how did you spot this approach in the first place? By 'feeling' the algebra? Thank you. @SarGe $\endgroup$ Aug 20, 2020 at 12:13
  • $\begingroup$ @IncredibleSimon, I think I have solved many such types of questions in olympiad worksheets. Maybe this was one of them. I don't remember exactly, though. :-) $\endgroup$
    – SarGe
    Aug 20, 2020 at 12:18
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Thanks to SarGe's hint, now I know how to solve it. I am posting below the rest part of a solution following SarGe's hint for future reference.

The question reduces to prove $(1-a)(1-b)(1-c)\ge0$. Assume the opposite. Then either $a,b,c\gt1$, or only one of $a,b,c$ is greater than 1 (say it is $a$). The former case is impossible because it contradicts $ab+bc+ac=1$ obviously. For the latter case, applying the triangle inequality, $b+c\gt a\gt1$, and then $ab+bc+ac=a(b+c)+bc\gt1$ which is a contradiction. Thus the proof is complete.

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  • $\begingroup$ That looks good. $\endgroup$ Aug 20, 2020 at 12:49
  • $\begingroup$ Good one... :-) $\endgroup$
    – SarGe
    Aug 20, 2020 at 12:50
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OK first let's expand the bracket

$(a+1)(b+1)(c+1)=abc+ab+ac+bc+a+b+c+1$.

Now we know that $ab+ac+bc=1$ so we actually need $abc+a+b+c+1 \leq 3$ or $abc+a+b+c \leq{2}$.

Since $a,b$ and $c$ form the sides of a triangle, we know that $a \leq b+c$ and $b \leq a+c$ and $c \leq a+b$.

I found it hard to progress from here and wondered if the result was actually true so did a thought experiment. Let us say $a,b$ and $c$ are all equal to $1/\sqrt{3}$. This would be an equilateral triangle and $ab+bc+ac=1/3+1/3+1/3=1$.

Then $(a+1)(b+1)(c+1)=abc+ab+ac+bc+a+b+c+1$=

$1/3 \sqrt{3}+1/3+1/3+1/3+1/\sqrt{3}+1/\sqrt{3}+1/\sqrt{3}+1=$

$1/3\sqrt{3}+1+\sqrt{3}+1$.

Which needs to be $\leq{4}$

Iff $1/3\sqrt{3} +\sqrt{3} \leq2$

iff $1/3+3 \leq 2\sqrt{3}$. Which is true.

Let's take another extreme case: $a$ and $b$ are just under $1$ and $c$ is close to $0$ then we can also have $ab+ac+bc=1$. Here $(a+1)(b+1)(c+1)$ will also be just under $4$ so I believe that the inequality is correct. I can show that we need $abc+a+b+c \leq{2}$ but don't know how to do that right now. I'll think about. But we haven't yet used the triangle inequalities so I suspect they are needed.

Not being able to finish it is killing me :)

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  • $\begingroup$ It is totally okay not to finish it, Simon. Great effort which is good for you! Also, your answer helps me to perceive and understand the question from a different angle. Thank you. :) @Simon Terrington $\endgroup$ Aug 20, 2020 at 12:09
  • $\begingroup$ Thanks IncredibleSimon; it's a pleasure and it looks like SarGe nailed it so I learned something too! :) $\endgroup$ Aug 20, 2020 at 12:17
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We need to prove $$abc+a+b+c\leq2$$ or $$(abc+(a+b+c)(ab+ac+bc))^2\leq4(ab+ac+bc)^3$$ or $$\prod_{cyc}(a(b+c-a)+bc)\geq0$$ and we are done!

We can get a last factoring by the following way.

For $ab+ac+bc=a^2$ we obtain: $$(abc+(a+b+c)(ab+ac+bc))^2=(abc+(a+b+c)a^2)^2=a^2(a^2+ab+ac+bc)^2=$$ $$=(ab+ac+bc)(2(ab+ac+bc))^2=4(ab+ac+bc)^3$$ and since we work with symmetric polynomials, we got the needed factorization.

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  • $\begingroup$ Hi Michael. This is truly exciting to see your ingenious approach. Do you mind clarifying a little bit more on how to get from the second inequality to the third? Thanks a lot. @Michael Rozenberg $\endgroup$ Aug 22, 2020 at 15:21
  • $\begingroup$ Hi Michael. I am sorry but I still do not get it. I recognized that what you added is to prove the equality conditions match. However, where does the final factorization come from exactly? I am really muddled up. Possibly there is some preliminary knowledge which I have not acquired. Sorry to trouble you again. Thank you. @Michael Rozenberg $\endgroup$ Aug 23, 2020 at 8:39
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    $\begingroup$ @IncredibleSimon I showed that if $ab+ac+bc=a^2$, so $(abc+(a+b+c)(ab+ac+bc))^2=4(ab+ac+bc)^3,$ which says that $4(ab+ac+bc)^3-(abc+(a+b+c)(ab+ac+bc))^2$ has a factor $ab+ac+bc-a^2$. Since our polynomial is symmetric we have also factors $ab+ac+bc-b^2$ and $ab+ac+bc-c^2.$ $\endgroup$ Aug 23, 2020 at 8:42
  • $\begingroup$ OMG, now I understand. Thank you so much, Michael. That is incredible. How did you spot this approach in the first place? It is definitely extraordinary. @Michael Rozenberg $\endgroup$ Aug 23, 2020 at 11:01

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