# Thread: BMO 2010/11 question on inequalities

1. ## BMO 2010/11 question on inequalities

Let $a, b$ and $c$ be lengths of the sides of a triangle. Suppose that $ab+bc+ca=1$. Show that $(a+1)(b+1)(c+1)<4$

2. ## Re: BMO 2010/11 question on inequalities

I was thinking something like this. Since each side isn't really defined to what it could be, and each side wouldn't exactly be the same length. We could say that C being the longest side will have a length of .25. C gets used twice, so you have .5. A will be .2, used it twice you get .4. Now you have a total of .9. Which leaves B to the shortest length getting .05, used twice will get you .1. With a total of 1. So now if we just plugin what we substituted earlier.
(.2+1)(.05+1)(.25+1) < 4
(1.2) (1.05) = 1.26 (1.25) = 1.575 < 4?

Though I'm not completely sure this is how it should be done, and if I made a mistake anywhere please feel free to correct me. This is how I would go about doing it though.

-Auri

3. ## Re: BMO 2010/11 question on inequalities

Originally Posted by Auri
I was thinking something like this. Since each side isn't really defined to what it could be, and each side wouldn't exactly be the same length. We could say that C being the longest side will have a length of .25. C gets used twice, so you have .5. A will be .2, used it twice you get .4. Now you have a total of .9. Which leaves B to the shortest length getting .05, used twice will get you .1. With a total of 1. So now if we just plugin what we substituted earlier.
(.2+1)(.05+1)(.25+1) < 4
(1.2) (1.05) = 1.26 (1.25) = 1.575 < 4?

Though I'm not completely sure this is how it should be done, and if I made a mistake anywhere please feel free to correct me. This is how I would go about doing it though.

-Auri
you have taken $a=0.2, b=0.05, c=0.25$. This doesn't satisfy $ab+bc+ca=1$. Moreover what you have attempted to do is called 'VERIFICATION'. This is not a proof.

4. ## Re: BMO 2010/11 question on inequalities

Oh, I'm so sorry! I read the problem completely wrong. I forgot it was ab, not A+B.

5. ## Re: BMO 2010/11 question on inequalities

I would rewrite:
$(a+1)\cdot(b+1)\cdot(c+1)<4$ as:
$(ab+a+b+1)(c+1)<4$
$\Leftrightarrow abc+ac+bc+c+ab+a+b+1<4$
$\Leftrightarrow abc+ac+bc+c+ab+a+b<3$ (1)

With the given $ab+bc+ca=1 \Leftrightarrow ab=1-bc-ca$(2)
Substituting (2) in (1) gives:
$abc+ac+bc+c+ab+a+b<3$
$\Leftrightarrow abc+ac+bc+c+1-bc-ac+a+b<3$
$\Leftrightarrow abc+a+b+c<2$
$\Leftrightarrow (abc)+(a+b+c)<2$

This looks very hard like if you write:
$ab+bc+ca=1 \Leftrightarrow ca+b(c+a)=1$
But I can't see any relation ...
But now I did nothing wit the fact a,b,c are the sides of a triangle. I don't know of this make sense, it was just an attempt

Do you have a hint or something? Or is the problem also not clear to you? ...

6. ## Re: BMO 2010/11 question on inequalities

Originally Posted by abhishekkgp
Let $a, b$ and $c$ be lengths of the sides of a triangle. Suppose that $ab+bc+ca=1$. Show that $(a+1)(b+1)(c+1)<4$
A little progress...

ab+bc+ac=1

1=b(a+c)+ac>b^2+ac

b^2+ac<1

0<ac<1-b^2=(1-b)(1+b)

b\in(0,1)

with the same argument like above one(?) can conclude that a,b,c\in(0,1)

7. ## Re: BMO 2010/11 question on inequalities

Originally Posted by abhishekkgp
Let $a, b$ and $c$ be lengths of the sides of a triangle. Suppose that $ab+bc+ca=1$. Show that $(a+1)(b+1)(c+1)<4$
Here's what I've tried.

We may assume without loss of generality that $a; from $ab+bc+ca=1$, we see then that $a<1$.

Sppose $b\leq1$. Since $a,b,c$ are the side lengths of a triangle, we have $b+a>c$ and then $1=c(b+a)+ab>c^2$; therefore $1>c$. From $a, it follows that $a,b,c$ are all less than $1$.

Now if $b>1$, then clearly $c>1$. This together with the above means that the product $N=(a-1)(b-1)(c-1)$ is negative.

Write $M=(a+1)(b+1)(c+1)=abc+(ab+ac+bc)+a+b+c+1=abc+a+b+c +2$; similarly, expand $N=(a-1)(b-1)(c-1)=abc+a+b+c-2$. We have $M-N=4$.

But $N<0$ which implies that $4=M-N>M$.

8. ## Re: BMO 2010/11 question on inequalities

As melese has pointed out, the key thing about a triangle is that each side must be shorter than the sum of the other two sides. So $b+c>a$. Write that as $a+b+c>2a$, and let $s = a+b+c$. Then $s-2a>0$, and similarly $s-2b>0$, $s-2c>0.$ Therefore

. . . . . . . . . . \begin{aligned}0&< (s-2a)(s-2b)(s-2c) \\ &= s^3 - 2s^2(a+b+c) + 4s(bc+ca+ab) -8abc\\ &= -s^3 + 4s - 8abc.\end{aligned}

Thus $8abc < 4s-s^3$, and so $8(s+abc) < 12s-s^3.$

However, if x>0 then $x^3-12x +16 = (x-2)^2(x+4) >0.$ It follows that $12s-s^3<16.$ Thus $8(s+abc) < 16$ and so $s+abc<2.$ As shown in Siron's comment above, that is equivalent to $(a+1)(b+1)(c+1)<4.$

9. ## Re: BMO 2010/11 question on inequalities

Originally Posted by melese
Here's what I've tried.

We may assume without loss of generality that $a; from $ab+bc+ca=1$, we see then that $a<1$.

Sppose $b\leq1$. Since $a,b,c$ are the side lengths of a triangle, we have $b+a>c$ and then $1=c(b+a)+ab>c^2$; therefore $1>c$. From $a, it follows that $a,b,c$ are all less than $1$.

Now if $b>1$, then clearly $c>1$. This together with the above means that the product $N=(a-1)(b-1)(c-1)$ is negative.

Write $M=(a+1)(b+1)(c+1)=abc+(ab+ac+bc)+a+b+c+1=abc+a+b+c +2$; similarly, expand $N=(a-1)(b-1)(c-1)=abc+a+b+c-2$. We have $M-N=4$.

But $N<0$ which implies that $4=M-N>M$.
Brilliant!! Opalg you too! Thanks both of you. Thank you Siron and Zarthustra.
EDIT: Hey melese. the part where you have assumed b>1 you say that c>1. but then ab+bc+ca=1 won't be satisfied.

thanks

11. ## Re: BMO 2010/11 question on inequalities

Originally Posted by abhishekkgp
Brilliant!! Opalg you too! Thanks both of you. Thank you Siron and Zarthustra.
EDIT: Hey melese. the part where you have assumed b>1 you say that c>1. but then ab+bc+ca=1 won't be satisfied.
You're right... I didn't notice. But I can regard this now as a contradiction,namely that $b>1$ is not possibe. Therefore, $ab+bc+ca=1$ only if $a,b,c$ are all less than $1$.

Do you think it's correct?

12. ## Re: BMO 2010/11 question on inequalities

Originally Posted by melese
You're right... I didn't notice. But I can regard this now as a contradiction,namely that $b>1$ is not possibe. Therefore, $ab+bc+ca=1$ only if $a,b,c$ are all less than $1$.

Do you think it's correct?
It is not evident to me how to prove $b \geq 1 \Rightarrow c \geq 1$. If we can prove this then it looks correct. Can you please elaborate on this point a bit more.

13. ## Re: BMO 2010/11 question on inequalities

Originally Posted by abhishekkgp
It is not evident to me how to prove $b \geq 1 \Rightarrow c \geq 1$. If we can prove this then it looks correct. Can you please elaborate on this point a bit more.
If $b>1$, then $c>1$. Simply because I assumed that $a without loss of generality.

14. ## Re: BMO 2010/11 question on inequalities

Originally Posted by melese
If $b>1$, then $c>1$. Simply because I assumed that $a without loss of generality.
i am so stupid.

15. ## Re: BMO 2010/11 question on inequalities

Originally Posted by melese
If $b>1$, then $c>1$. Simply because I assumed that $a without loss of generality.
However, it's still not clear to me how you get that all three have to be less than one. Let's say you assume a < b < c. Then you can easily show that a and b must be less than one. Why does c have to be less than one? And if you assume a different ordering, well, that's a separate case. In each possible ordering, you still get one side that could, theoretically, be greater than one. The fact that they're disjoint cases is neither here nor there.

Incidentally, I think it's more realistic to allow equalities in your assumption. That is, you really should assume $a\le b\le c.$

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