$(a+1)(b+1)(c+1)\leq4$ for triangle sides $a,b,c$ with $ab+bc+ac=1$

Question

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!

Answer 1

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 $$\prod_{cyc}(1+a)+\prod_{cyc}(1-a)=4$$

Answer 2

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 :)

Answer 3

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.

Answer 4

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.