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Math Help - Proving inequalities

  1. #1
    Junior Member Kaloda's Avatar
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    Proving inequalities

    Let a,b, c be non-negative real nos. such that a+b+c=1. Prove that a2+b2+c2+sqrt(12abc)<=1.

    Please help me with this one. I don't really have an idea how to prove inequalities. TY.
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  2. #2
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    Re: Proving inequalities

    Given a+b+c = 1, squaring both sides yields a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 1.

    Here, if we can show that 2ab + 2bc + 2ca \ge \sqrt{12abc} or ab + bc + ca \ge \sqrt{3abc}, we are done.

    Squaring both sides of this inequality, a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a+b+c) \ge 3abc \Rightarrow a^2b^2 + b^2c^2 + c^2a^2 \ge abc (by replacing a+b+c with 1). This is what we want to prove.


    By the Cauchy-Schwarz inequality, (a^2b^2 + b^2c^2 + c^2a^2)(1+1+1) \ge (ab + bc + ca)^2. Simplifying, we obtain

    2(a^2b^2 + b^2c^2 + c^2a^2) \ge 2abc(a+b+c) \Rightarrow a^2b^2 + b^2c^2 + c^2a^2 \ge abc, which is exactly what we wanted to prove. Hence we're done.
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  3. #3
    Junior Member Kaloda's Avatar
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    Re: Proving inequalities

    Thank you. But is there another way of proving this without using that Cauchy-Schwarz inequality because I think we didn't tackled it in our class discussion yet.
    Or can you give me the brief proof of that inequality without using calculus because I don't have a single knowledge in calculus. Please. TY very MUCH.
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  4. #4
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    Re: Proving inequalities

    To prove inequality (a^2b^2+a^2c^2+b^2c^2)(1+1+1) \geq (ab+ac+bc)^2
    consider quadratic function
    (abx+1)^2+(acx+1)^2+(bcx+1)^2
    Since it is sum of squares than it is never negative, therefore its discriminant is never positive: D \leq 0.

    Now expand function (recall that (x+y)^2=x^2+2xy+y^2): a^2b^2x^2+2abx+1+a^2c^2x^2+2acx+1+b^2c^2x^2+2bcx+1
    or
    (a^2b^2+a^2c^2+b^2c^2)x^2+2(ab+ac+bc)x+1+1+1

    Find discriminant: D=(ab+ac+bc)^2- (1+1+1)(a^2b^2+a^2c^2+b^2c^2)
    Also, D \leq 0
    Therefore
    (ab+ac+bc)^2- (1+1+1)(a^2b^2+a^2c^2+b^2c^2) \leq 0
    Or

     (1+1+1)(a^2b^2+a^2c^2+b^2c^2) \geq (ab+ac+bc)^2

    As you see no calculus needed.
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  5. #5
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    Re: Proving inequalities

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  6. #6
    Junior Member Kaloda's Avatar
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    Re: Proving inequalities

    Why does the discriminant is non-positive? What I knew is that the root of a quadratic function will become imaginary if the discriminant is negative.

    Do you know any OTHER WAY or SOLUTION to approach this problem? I think your solutions combined is very long and complicated. But thank you very very much for your replies.

    Sorry for my ignorance in higher math. I'm usually good at math but not on this one. It's my first time to solve a problem like this.
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  7. #7
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    Re: Proving inequalities

    That's the way inequality problems are; usually you have to use some inequality such as Cauchy-Schwarz, AM-GM, Jensen's, rearrangement inequality, or Muirhead's. Usually you don't see such inequality problems until you take exams such as the USAMO or IMO, which are pretty difficult.

    Also, do you see how I got from (a^2b^2 + b^2c^2 + c^2a^2)(1+1+1) \ge (ab + bc + ca)^2 to a^2b^2 + b^2c^2 + c^2a^2 \ge abc? Sorry I skipped a few steps...but if you expand the RHS you should obtain the same thing...
    Last edited by richard1234; June 16th 2012 at 01:42 PM.
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  8. #8
    Junior Member Kaloda's Avatar
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    Re: Proving inequalities

    Yea. I think I did traced how you came into that equation.
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  9. #9
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    Re: Proving inequalities

    Okay, great. Yeah, the (a^2b^2 + b^2c^2 + c^2a^2)(1+1+1) \ge (ab + bc + ca)^2 is essentially the Cauchy-Schwarz inequality.


    In general, the Cauchy-Schwarz inequality says that, for non-negative numbers a_1, a_2, ..., a_n and b_1, b_2, ..., b_n, that


    (a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_2^2 + ... + b_n^2) \ge (a_1b_1 + a_2b_2 + ... + a_nb_n)^2


    Equality occurs if and only if a_i is a constant multiple of b_i, that is, a_i = kb_i for all i. It can be generalized even further to complex numbers and vectors.
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