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Math Help - Inequality

  1. #1
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    Inequality

    1. Prove that if a;b\in\mathbb{R}^+ and a+b=1, then
      \left(a+\frac 1a\right)^2+\left(b+\frac 1b\right)^2\ge \frac {25}2.
    2. Similarly to the previous problem:
      Prove that if a;b;c\in\mathbb{R}^+ and a+b+c=1, then
      \left(a+\frac 1a\right)^2+\left(b+\frac 1b\right)^2+\left(c+\frac 1c\right)^2\ge \frac {100}3.
    3. Let's generalize the problems above:
      Prove that if a_1;a_2;\ldots ;a_n\in\mathbb{R}^+, n\ge 2, n\in\mathbb{N}, and a_1+a_2+\ldots +a_n=1 then
      \left(a_1+\frac 1{a_1}\right)^2+\left(a_2+\frac 1{a_2}\right)^2+\ldots +\left(a_n+\frac 1{a_n}\right)^2\ge K.
      Find the maximum possible value of K.
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by james_bond View Post
    1. Prove that if a;b\in\mathbb{R}^+ and a+b=1, then
      \left(a+\frac 1a\right)^2+\left(b+\frac 1b\right)^2\ge \frac {25}2.
    Equality for a=b=\frac12
    By Cauchy-Schwarz we have  (a+\frac 1a)^2+(b+\frac 1b)^2 \geq \frac{(a+\frac 1a + b+\frac 1b)^2}{1 + 1} = \frac{(1+\frac{1}{ab})^2}{2}

    Since (a+b)^2 = 1 \geq 4ab \Rightarrow \frac{(1 + \frac1{ab})^2}{2} \geq \frac{25}{2}.

    Thus we have our solution
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  3. #3
    Lord of certain Rings
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    I post the remaining separately because a small trick there, wont work any longer

    I will directly attack the generalization,

    Claim:  K = \frac{(n^2 + 1)^2}{n}
    Proof:
    Denote LHS of your inequality by I,
    AM-HM says:
    (\sum a_i)(\sum \frac1{a_i}) \geq n^2 \Rightarrow (1 + \sum \frac1{a_i})^2 \geq (n^2 + 1)^2-------------(*)
    Applying CS:
    I \geq \frac{{(a_1+\frac 1{a_1} + a_2+\frac 1{a_2} + \ldots a_n+\frac 1{a_n}})^2}{1+1+1 \ldots +1}
    But ({a_1+\frac 1{a_1} + a_2+\frac 1{a_2} + \ldots a_n+\frac 1{a_n}} = 1 + \sum \frac1{a_i}
    So I \geq \frac{{(1 + \sum \frac1{a_i})}^2}{n}.
    But by (*), \frac{{(1 + \sum \frac1{a_i})}^2}{n}\geq \frac{(n^2 + 1)^2}{n}
    Thus I \geq \frac{(n^2 + 1)^2}{n}
    Equality for all a_is equal.
    K cant get any better than that because we showed equality achieving
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