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Math Help - Inequality

  1. #1
    Junior Member
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    Mar 2010
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    Inequality

    if a,b,c,d>0 and

    a+b+c+d=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{  1}{d}

    prove that

    2(a+b+c+d)\geq\sqrt{a^{2}+3}+\sqrt{b^{2}+3}+\sqrt{  c^{2}+3}+\sqrt{d^{2}+3}
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  2. #2
    Super Member
    Joined
    Jan 2009
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    Using Cachy Inequality , we have

     \sqrt{ \left( \sqrt{a}^2 + \sqrt{b}^2 + \sqrt{c}^2 + \sqrt{d}^2 \right) \left( \sqrt{ \frac{a^2+3}{a} }^2 + \sqrt{ \frac{b^2+3}{b} }^2 + \sqrt{ \frac{c^2+3}{c} }^2 + \sqrt{ \frac{d^2+3}{d} }^2 \right) } \geq \sqrt{a^2 +3} + \sqrt{b^2 +3} + \sqrt{c^2 +3} + \sqrt{d^2 +3}

     \sqrt{ ( a+b+c+d) [ a + b + c + d + 3( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} ) ] } \geq \sqrt{a^2 +3} + \sqrt{b^2 +3} + \sqrt{c^2 +3} + \sqrt{d^2 +3}

     \sqrt{ ( a+b+c+d) [ a + b + c + d + 3(a+b+c+d ) ] } \geq \sqrt{a^2 +3} + \sqrt{b^2 +3} + \sqrt{c^2 +3} + \sqrt{d^2 +3}

     \sqrt{4 (a+b+c+d)^2 } \geq \sqrt{a^2 +3} + \sqrt{b^2 +3} + \sqrt{c^2 +3} + \sqrt{d^2 +3}

     2(a+b+c+d) \geq \sqrt{a^2 +3} + \sqrt{b^2 +3} + \sqrt{c^2 +3} + \sqrt{d^2 +3}
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  3. #3
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    Thank u
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