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Math Help - Secondary school's competition - marathon.

  1. #1
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    Secondary school's competition - marathon.

    Hello everyone.
    I would like to start 'marathon' of the exercises from different competition in the secondary school level. (ages 15-18)

    let's start with inequality
    Exercise 1.
    Let a,b be positive real numbers. Prove that
    \left(1+\frac{a}{b} \right)^m+\left(1+\frac{b}{a} \right)^m \ge 2^{m+1}
    where m is natural.

    Enjoy!
    Last edited by ghostbuster9; August 29th 2010 at 03:25 AM.
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  2. #2
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    If a=b, we have the equality so assume, wlog, that a>b and that a=kb, \ \  k>1.

    We then have to prove that

    (1+k)^{m}+(1+1/k)^{m} > 2^{m+1}.

    Now, [(1+k)^{m/2}-(1+1/k)^{m/2}]^{2}=(1+k)^{m}+(1+1/k)^{m}-2(1+k)^{m/2}(1+1/k)^{m/2}>0,
    so,
    (1+k)^{m}+(1+1/k)^{m}>2(1+k)^{m/2}(1+1/k)^{m/2}=2[(1+k)(1+1/k)]^{m/2}

    =2(2+k+1/k)^{m/2}.

    If k>2, then 2+k+1/k>4+1/k=2^{2}+1/k and the result follows, so assume that 1<k<2 and that k=1+\alpha with 0<\alpha<1.

    Then,
    2+k+1/k=2+(1+\alpha)+1/(1+\alpha)=3+\alpha+1-\alpha+\alpha^{2}-...,

    =4+\alpha^{2}-\alpha^{3}+...>2^{2}, and again the result follows.
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  3. #3
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    You could a bit easier: AM-GM two times. But your solution is also OK. Now is your turn - send your exercise.
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  4. #4
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    O.K.

    Exercise 2.

    Prove that there do not exist positive integers x, y, z such that

    x^2 +y^2 + z^2 = 2xyz.
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  5. #5
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    x^2 +y^2 + z^2 = 2xyz
    (x+y+z)^2=2(x+1)(y+1)(z+1)-2(x+y+z)-2
    (x+y+z)^2+2(x+y+z)=2(x+1)(y+1)(z+1)-2
    (x+y+z)(x+y+z+2)=2[(x+1)(y+1)(z+1)-1]
    what is impossible.

    Exercise 3.
    Calculate
    \sqrt{ \underbrace{44...4}_{2n}+\underbrace{22...2}_{n+1}  +\underbrace{88...8}_{n}+7}
    Last edited by ghostbuster9; August 30th 2010 at 06:40 AM.
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