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Math Help - Inequality

  1. #1
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    Inequality

    Prove that if ab+bc+ca=\frac13 ( a,b,c are positive reals) then
    \frac a{a^2-bc+1}+\frac b{b^2-ca+1}+\frac c{c^2-ab+1}\ge \frac 1{a+b+c}.
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  2. #2
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    I am sure if you multiplied the inequality by the LCD then that may help.
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  3. #3
    Senior Member JaneBennet's Avatar
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    This problem looks similar to this one. Is the method to be used similar?
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  4. #4
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    Probably. (I haven't done it so don't know but it's from a series...) Maybe the inequality could be used as we proved it.
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  5. #5
    Senior Member JaneBennet's Avatar
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    This problem has been gripping me for a while now and still I canít solve it. Hereís what Iíve done with the inequality so far. Multiply through by a+b+c:

    \sum{\frac{a^2+ab+ca}{a^2-bc+1}}\ \geq\ 1

    \sum{\frac{a^2-bc+\frac{1}{3}}{a^2-bc+1}}\ \geq\ 1

    \sum{\left[1-\frac{2}{3}\,\left(\frac{1}{a^2-bc+1}\right)\right]}\ \geq\ 1

    \therefore\ \frac{1}{a^2-bc+1}+\frac{1}{b^2-ca+1}+\frac{1}{c^2-ab+1}\ \leq\ 3

    Now

    \frac{1}{a^2-bc+1}\ =\ \frac{1}{a^2+ab+ca+\frac{2}{3}}\ =\ \frac{3}{3a(a+b+c)+2}

    Thus

    \frac{1}{3a(a+b+c)+2}+\frac{1}{3b(a+b+c)+2}+\frac{  1}{3c(a+b+c)+2}\ \leq\ 1

    Let A=\sqrt{3}a,\ B=\sqrt{3}b,\ C=\sqrt{3}c. Then AB+BC+CA=1 and the inequality becomes


    \color{white}.\hspace{15mm}\color{black}\frac{1}{A  (A+B+C)+2}+\frac{1}{B(A+B+C)+2}+\frac{1}{C(A+B+C)+  2}\ \leq\ 1


    This is what needs to be proved. I think the trick is to let A=\tan{\alpha},\ B=\tan{\beta},\ C=\tan{\gamma}, where \alpha+\beta+\gamma=\frac{\pi}{2}, but I still canít do it.
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  6. #6
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    This problem has also got my head interacting with brick walls.

    Following JaneBennet's ideas, we want to show that if A, B, C are positive numbers with BC+CA+AB=1, then

    \color{white}.\hspace{15mm}\color{black}\frac{1}{A  (A+B+C)+2} + \frac{1}{B(A+B+C)+2} + \frac{1}{C(A+B+C)+2}\ \leq\ 1.

    Let x^3 - px^2 + qx -r = 0
    be the cubic equation with roots A, B and C. Then p=A+B+C, q=BC+CA+AB and r=ABC. Hence q=1.

    The equation (x/p)^3 - p(x/p)^2 + (x/p) - r = 0 has roots Ap, Bp and Cp. In other words, the equation  x^3 - p^2x^2 + p^2x - p^3r = 0 has roots A(A+B+C), B(A+B+C), C(A+B+C).

    Therefore the equation  (x-2)^3 - p^2(x-2)^2 + p^2(x-2) - p^3r = 0 has roots A(A+B+C)+2, B(A+B+C)+2, C(A+B+C)+2. In other words (multiplying out the brackets), the equation x^3 - (p^2+6)x^2 + (5p^2+12)x - (p^3r+6p^2+8) = 0 has those roots.

    Replacing x by 1/x, we see that the equation (p^3r+6p^2+8)x^3 - (5p^2+12)x^2 + (p^2+6)x - 1 = 0 has roots \frac1{A(A+B+C)+2}, \frac1{B(A+B+C)+2}, \frac1{C(A+B+C)+2}.

    The sum of the roots of this cubic equation is also the coefficient of x^2 divided by the coefficient of x^3. So to prove our inequality we need to show that \frac{5p^2+12}{p^3r+6p^2+8}\leqslant1, where p=A+B+C and r=ABC. An equivalent inequality is (rp+1)p^2\geqslant4.

    Now what?
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  7. #7
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    Quote Originally Posted by JaneBennet View Post
    This problem has been gripping me for a while now and still I canít solve it. Hereís what Iíve done with the inequality so far. Multiply through by a+b+c:

    \sum{\frac{a^2+ab+ca}{a^2-bc+1}}\ \geq\ 1

    \sum{\frac{a^2-bc+\frac{1}{3}}{a^2-bc+1}}\ \geq\ 1

    \sum{\left[1-\frac{2}{3}\,\left(\frac{1}{a^2-bc+1}\right)\right]}\ \geq\ 1

    \therefore\ \frac{1}{a^2-bc+1}+\frac{1}{b^2-ca+1}+\frac{1}{c^2-ab+1}\ \leq\ 3

    Now

    \frac{1}{a^2-bc+1}\ =\ \frac{1}{a^2+ab+ca+\frac{2}{3}}\ =\ \frac{3}{3a(a+b+c)+2}

    Thus

    \frac{1}{3a(a+b+c)+2}+\frac{1}{3b(a+b+c)+2}+\frac{  1}{3c(a+b+c)+2}\ \leq\ 1

    I can show a+b+c > 1 ,and hence we will be required to show,
    \frac{1}{3a+2}+\frac{1}{3b+2}+\frac{1}{3c+2}\ \leq\ 1
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  8. #8
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    Quote Originally Posted by Isomorphism View Post
    I can show a+b+c > 1 ,and hence we will be required to show,
    \frac{1}{3a+2}+\frac{1}{3b+2}+\frac{1}{3c+2}\ \leq\ 1
    Unfortunately that inequality need not hold. For example, if a=0, b=c=1/√3, then bc+ca+ab=1, but \frac{1}{3a+2}+\frac{1}{3b+2}+\frac{1}{3c+2}>1.03.
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  9. #9
    Lord of certain Rings
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    Quote Originally Posted by Opalg View Post
    Unfortunately that inequality need not hold. For example, if a=0, b=c=1/√3, then bc+ca+ab=1, but \frac{1}{3a+2}+\frac{1}{3b+2}+\frac{1}{3c+2}>1.03.
    ya thats true, that means it's a strict inequality. Hmm i am trying it from a long time, but no use
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  10. #10
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    Okay, I can complete the proof of this, but it's not at all illuminating. There must be a more conceptual explanation for this strange inequality.

    The situation is that A, B, C are positive numbers with BC+CA+AB=1. If p=A+B+C and r=ABC then we want to show that rp^3+p^2\geqslant4. Since the proof involves calculus, I'll change the notation (again!) and write x, y, z for A, B, C.

    So x, y, z are positive reals with yz+zx+xy=1, and we want to show that f(x,y,z) := xyz(x+y+z)^3 + (x+y+z)^2\geqslant4. Let D=\{(x,y,z)\in\mathbb{R}^3:x,y,z\geqslant0, yz+zx+xy=1\}. Then we want to show that the minimum value of f on D is 4.

    The boundary of D consists of points for which one of the coordinates is zero. Then the other two coordinates must be w and 1/w, for some positive number w. The value of f at this point is (w+w^{-1})^2, and it is easy to see that the minimum of this expression is 4 (occurring when w=1).

    To find the extrema of f in the interior of D, use the method of Lagrange multipliers, and find the zeros of the partial derivatives of g(x,y,z)=f(x,y,z)+λ(yz+zx+xy-1):
    \frac{\partial g}{\partial x} = yz(x+y+z)^3 + 3xyz(x+y+z)^2 +2(x+y+z) + \lambda(y+z) = 0 ,
    \frac{\partial g}{\partial y} = zx(x+y+z)^3 + 3xyz(x+y+z)^2 +2(x+y+z) + \lambda(z+x) = 0 ,
    \frac{\partial g}{\partial z} = xy(x+y+z)^3 + 3xyz(x+y+z)^2 +2(x+y+z) + \lambda(x+y) = 0 .

    Subtract the last of these from the middle one, to get (z-y)\{x(x+y+z)^3 + \lambda\} = 0 (and two similar equations with x,y and z cyclically interchanged). If x,y and z are all different then \lambda = -x(x+y+z)^3 = -y(x+y+z)^3 = -z(x+y+z)^3, which is impossible (since it implies that x, y and z are all equal).

    Therefore the extreme values of f on D must occur at points where at least two of x, y and z are equal. Suppose for example that z=y. Then the conditon (x,y,z)∈D becomes y(2x+y)=1, and so x={\textstyle\frac12}(y^{-1}-y). Note that this implies 0<y\leqslant1 (otherwise x would be negative). Also, x+y+y = \frac{1+3y^2}{2y} and xy^2 = {\textstyle\frac12}y(1-y^2).

    Then f(x,y,y) = {\textstyle\frac12}y(1-y^2)\frac{(1+3y^2)^3}{(2y)^3} + \frac{(1+3y^2)^2}{(2y)^2}, which simplifies(?) to \frac{5+32y^2+54y^4-27y^8}{16y^2}. We want this to be greater than or equal to 4, or in other words 5+32y^2+54y^4-27y^8\geqslant64y^2. When the term 64y^2 is taken over to the left side, this factorises as (1-y^2)(1-3y^2)^2(5+3y^2)\geqslant0, and this is clearly true when 0<y≤1.
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  11. #11
    Senior Member JaneBennet's Avatar
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    Actually, it can be shown that p^2\geq3. By the Cauchy–Schwarz inequality,

    \sqrt{A^2+B^2+C^2}\sqrt{B^2+C^2+A^2}\ \geq\ AB+BC+CA\

    \Rightarrow\ A^2+B^2+C^2 \geq\ 1

    \Rightarrow\ (A+B+C)^2-2(AB+BC+CA)\ \geq\ 1

    \Rightarrow\ (A+B+C)^2\ \geq\ 1+2\ =\ 3

    So it would appear to be simpler to prove that rp^3\geq1. The trouble is that r\ngeq\frac{1}{3\sqrt{3}}; in fact, the inequality is the other way round: ABC\leq\frac{1}{3\sqrt{3}}.
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  12. #12
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    Quote Originally Posted by JaneBennet View Post
    So it would appear to be simpler to prove that rp^3\geq1. The trouble is that r\ngeq\frac{1}{3\sqrt{3}}; in fact, the inequality is the other way round: ABC\leq\frac{1}{3\sqrt{3}}.
    That's exactly what makes it such a tough problem. In fact, r can be 0, so it really doesn't help to know that p^2\geq3.
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  13. #13
    Senior Member JaneBennet's Avatar
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    r canít be 0 because a, b, and c (and hence A, B, and C) are all positive.
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  14. #14
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    Quote Originally Posted by JaneBennet View Post
    r canít be 0 because a, b, and c (and hence A, B, and C) are all positive.
    r can be arbitrarily close to zero (if for example a is very small, and b and c are close to 1/√3), so my previous comment still holds good.
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  15. #15
    Senior Member JaneBennet's Avatar
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    By the way, some people think this problem is easy.

    Art of Problem Solving Forum

    Unfortunately I donít understand the steps in their solution.
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