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Thread: BMO 2010/11 question on inequalities

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    Senior Member abhishekkgp's Avatar
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    BMO 2010/11 question on inequalities

    Let $\displaystyle a, b$ and $\displaystyle c$ be lengths of the sides of a triangle. Suppose that $\displaystyle ab+bc+ca=1$. Show that $\displaystyle (a+1)(b+1)(c+1)<4$
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    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
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    Senior Member abhishekkgp's Avatar
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    Re: BMO 2010/11 question on inequalities

    Quote Originally Posted by Auri View Post
    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 $\displaystyle a=0.2, b=0.05, c=0.25$. This doesn't satisfy $\displaystyle ab+bc+ca=1$. Moreover what you have attempted to do is called 'VERIFICATION'. This is not a proof.
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    Newbie Auri's Avatar
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    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.
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    MHF Contributor Siron's Avatar
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    Re: BMO 2010/11 question on inequalities

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

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

    This looks very hard like if you write:
    $\displaystyle ab+bc+ca=1 \Leftrightarrow ca+b(c+a)=1$
    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? ...
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: BMO 2010/11 question on inequalities

    Quote Originally Posted by abhishekkgp View Post
    Let $\displaystyle a, b$ and $\displaystyle c$ be lengths of the sides of a triangle. Suppose that $\displaystyle ab+bc+ca=1$. Show that $\displaystyle (a+1)(b+1)(c+1)<4$
    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)
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    Re: BMO 2010/11 question on inequalities

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

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

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

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

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

    But $\displaystyle N<0$ which implies that $\displaystyle 4=M-N>M$.
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    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 $\displaystyle b+c>a$. Write that as $\displaystyle a+b+c>2a$, and let $\displaystyle s = a+b+c$. Then $\displaystyle s-2a>0$, and similarly $\displaystyle s-2b>0$, $\displaystyle s-2c>0.$ Therefore

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

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

    However, if x>0 then $\displaystyle x^3-12x +16 = (x-2)^2(x+4) >0.$ It follows that $\displaystyle 12s-s^3<16.$ Thus $\displaystyle 8(s+abc) < 16$ and so $\displaystyle s+abc<2.$ As shown in Siron's comment above, that is equivalent to $\displaystyle (a+1)(b+1)(c+1)<4.$
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    Re: BMO 2010/11 question on inequalities

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

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

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

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

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

    But $\displaystyle N<0$ which implies that $\displaystyle 4=M-N>M$.
    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.
    Last edited by abhishekkgp; Aug 4th 2011 at 07:31 AM.
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    Re: BMO 2010/11 question on inequalities

    thanks
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    Re: BMO 2010/11 question on inequalities

    Quote Originally Posted by abhishekkgp View Post
    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 $\displaystyle b>1$ is not possibe. Therefore, $\displaystyle ab+bc+ca=1$ only if $\displaystyle a,b,c$ are all less than $\displaystyle 1$.

    Do you think it's correct?
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    Re: BMO 2010/11 question on inequalities

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

    Do you think it's correct?
    It is not evident to me how to prove $\displaystyle b \geq 1 \Rightarrow c \geq 1$. If we can prove this then it looks correct. Can you please elaborate on this point a bit more.
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    Re: BMO 2010/11 question on inequalities

    Quote Originally Posted by abhishekkgp View Post
    It is not evident to me how to prove $\displaystyle b \geq 1 \Rightarrow c \geq 1$. If we can prove this then it looks correct. Can you please elaborate on this point a bit more.
    If $\displaystyle b>1$, then $\displaystyle c>1$. Simply because I assumed that $\displaystyle a<b<c$ without loss of generality.
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    Senior Member abhishekkgp's Avatar
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    Re: BMO 2010/11 question on inequalities

    Quote Originally Posted by melese View Post
    If $\displaystyle b>1$, then $\displaystyle c>1$. Simply because I assumed that $\displaystyle a<b<c$ without loss of generality.
    i am so stupid.
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    Re: BMO 2010/11 question on inequalities

    Quote Originally Posted by melese View Post
    If $\displaystyle b>1$, then $\displaystyle c>1$. Simply because I assumed that $\displaystyle a<b<c$ 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 $\displaystyle a\le b\le c.$
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