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Math Help - Prove

  1. #1
    Super Member dhiab's Avatar
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    Prove

    a_1 ,a_2 ,a_3 <br /> <br />

    is the positifs reals numbers, Prove :
    3\left( {\frac{1}{{a_1 a_2 }} + \frac{1}{{a_3 a_2 }} + \frac{1}{{a_1 a_3 }}} \right) \ge 4\left( {\frac{1}{{a_1 + a_2 }} + \frac{1}{{a_3 + a_2 }} + \frac{1}{{a_1 + a_3 }}} \right)^2
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    First note that by AM–GM, \frac{(a_i+a_j)^2}4\ge a_ia_j and so \frac{a_ia_j}{(a_i+a_j)^2}\le\frac14.

    Hence

    \left(\frac{1}{{a_1 + a_2 }} + \frac{1}{{a_2 + a_3 }} + \frac{1}{{a_3 + a_1 }}\right)^2


    =\ \left(\frac{\sqrt{a_1a_2}}{(a_1 + a_2)\sqrt{a_1a_2}} + \frac{\sqrt{a_2a_3}}{(a_2 + a_3)\sqrt{a_2a_3}} + \frac{\sqrt{a_3a_1}}{(a_3 + a_1)\sqrt{a_3a_1}}\right)^2


    \leqslant\ \left(\frac{a_1a_2}{(a_1+a_2)^2}+\frac{a_2a_3}{(a_  2+a_3)^2}+\frac{a_3a_1}{(a_3+a_1)^2}\right)\left(\  frac1{a_1a_2}+\frac1{a_2a_3}+\frac1{a_3a_1}\right) (Cauchy–Schwarz)


    \leqslant\ \frac34\left(\frac1{a_1a_2}+\frac1{a_2a_3}+\frac1{  a_3a_1}\right) (by above)
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