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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestUsing T2's Lemma,

$\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a} = \dfrac{\left(\sqrt{2}\right)^2}{a+b}+\dfrac{\left(\sqrt{2}\right)^2}{b+c}+\dfrac{\left(\sqrt{2}\right)^2}{c+a} \geq \dfrac{\left(3\sqrt{2}\right)^2}{2(a+b+c)} = \dfrac{18}{2(a+b+c)}= \dfrac{9}{a+b+c} \ \Huge{\ddot\smile}$

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HAAHAHHA T2!

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Same method.

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Nice!

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I have a same way.

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Do AM-HM on the numbers $a+b,b+c,c+a$.

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T2 is a cuter method though :P

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From your inequality, we have: $2(a+b+c)(\frac { 1 }{ a+b } +\frac { 1 }{ b+c } +\frac { 1 }{ c+a } )\ge 9$

1, We can use AM-GM for 3 numbers in left-hand side . 2, We can extract in left-hand side and use AM-GM for 2 numbers too ^_^

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can someone tell me what is T2's Lemma?

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Titu's Lemma

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Thank you ... how do u do by AM-HM inequality?

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Power Mean (QAGH)

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