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Thread: Inequality

  1. #1
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    Inequality

    Prove that if $\displaystyle ab + ac + bc = 3$:
    $\displaystyle 8(a + b + c)^2\geq9(a + b)(a + c)(b + c)$
    ($\displaystyle a,b,c\ge 0$)
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  2. #2
    Super Member wingless's Avatar
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    The question says $\displaystyle a,b,c\ge 0$..
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  3. #3
    Senior Member JaneBennet's Avatar
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    Suppose one of $\displaystyle a,b,c$ is 0. Then the other two can’t be 0 if the condition $\displaystyle ab+bc+ca=3$ is to hold. Say, we have $\displaystyle a=0$. Then $\displaystyle bc=3$ and the inequality becomes

    $\displaystyle \color{white}.\quad.$ $\displaystyle 8\left(b+\frac{3}{b}\right)^2\geq9bc\left(b+\frac{ 3}{b}\right)=27\left(b+\frac{3}{b}\right)$

    $\displaystyle \Leftrightarrow\ 8\left(b+\frac{3}{b}\right)\geq27$

    $\displaystyle \Leftrightarrow\ 8b^2-27b+24\geq0$

    The LHS, as a quadratic expression in $\displaystyle b$, has negative discriminant and so the inequality is true.

    So now we’ll assume $\displaystyle a,b,c>0$. Let $\displaystyle a=\sqrt{3}\,A,\ b=\sqrt{3}\,B,\ c=\sqrt{3}\,C$. Then $\displaystyle AB+BC+CA=1$ and the inequality becomes

    $\displaystyle 8(A+B+C)^2\geq9\sqrt{3}(A+B)(B+C)(C+A)$

    Since $\displaystyle A,B,C$ are positive and $\displaystyle AB+BC+CA=1$, we can make the following substitutions:

    $\displaystyle A=\tan{\alpha},\ B=\tan{\beta},\ C=\tan{\gamma}$

    where $\displaystyle \alpha+\beta+\gamma=\frac{\pi}{2}$. Then observe that

    $\displaystyle A+B=\frac{\sin{\alpha}}{\cos{\alpha}}+\frac{\sin{\ beta}}{\cos{\beta}}$

    $\displaystyle \color{white}.\hspace{10mm}.$ $\displaystyle =\frac{\sin{\alpha}\cos{\beta}+\cos{\alpha}\sin{\b eta}}{\cos{\alpha}\cos{\beta}}$

    $\displaystyle \color{white}.\hspace{10mm}.$ $\displaystyle =\frac{\sin{\left(\alpha+\beta\right)}}{\cos{\alph a}\cos{\beta}}$

    $\displaystyle \color{white}.\hspace{10mm}.$ $\displaystyle =\frac{\cos{\left(\frac{\pi}{2}-\left[\alpha+\beta\right]\right)}}{\cos{\alpha}\cos{\beta}}$

    $\displaystyle \color{white}.\hspace{10mm}.$ $\displaystyle =\frac{\cos{\gamma}}{\cos{\alpha}\cos{\beta}}$

    Hence

    $\displaystyle (A+B)(B+C)(C+A)$ $\displaystyle =$ $\displaystyle \left(\frac{\cos{\gamma}}{\cos{\alpha}\cos{\beta}} \right)\left(\frac{\cos{\alpha}}{\cos{\beta}\cos{\ gamma}}\right)\left(\frac{\cos{\beta}}{\cos{\gamma }\cos{\alpha}}\right)=\frac{1}{\cos{\alpha}\cos{\b eta}\cos{\gamma}}$

    So it comes down to proving this inequality:

    $\displaystyle \cos{\alpha}\cos{\beta}\cos{\gamma}\left(\tan{\alp ha}+\tan{\beta}+\tan{\gamma}\right)^2\ \geq\ \frac{9\sqrt{3}}{8}$
    Last edited by JaneBennet; Feb 22nd 2008 at 12:36 PM.
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  4. #4
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    hmmm... don’t know why I was heavily neg repped on this question =\ but yer .. anyways...

    ^^ top was interesting

    After about 2 pages of working trying to find similarities between the two I got this...


    LHS

    8(a+b+c)^2

    8a^2 + 16ab + 16ac + 8b^2 + 16bc + 8c^2

    8(a^2 + b^2 + c^2) + 16(ac + ab + cd)

    8(a^2 + b^2 + c^2) + 16(3)

    8(a^2 + b^2 + c^2) + 48

    8(a^2 + b^2 + c^2 + 6)

    RHS

    9(a+b)(a+c)(b+c)

    9a^2b + 9a^2c + 9ab^2 + 18abc + 9ac^2 + 9b^2c + 9bc^2

    9a(ab + ac) + 9b(ab + bc) + 9c(ac + bc) +18(abc)

    9a(3-bc) + 9b(3-ac) + 9c(3-ab) +18abc

    27a – 9abc + 27b – 9abc + 27c -9abc + 18abc

    27(a + b + c) – 9abc

    9(3(a + b + c) - abc)

    So we can say….

    8(a^2 + b^2 + c^2 + 6) => 9(3(a + b + c) - abc)

    when ab + cb + bc = 3…..

    I also noticed a + b + c appears In the RHS which also appears in the LHS

    So we can let a + b + c = d

    And say:

    8d^2 => 27d – 9abc

    still doesn’t help unless we know a b or c = 0…
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