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Prove that,\[\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a} \geq \dfrac{9}{a+b+c}\\a,b,c \in R^+\]

Note by Adarsh Kumar 3 years ago

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Using 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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I have a same way.

Nice!

Same method.

HAAHAHHA T2!

Do AM-HM on the numbers \(a+b,b+c,c+a\).

T2 is a cuter method though :P

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 ^_^

can someone tell me what is T2's Lemma?

Titu's Lemma

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

@Ganesh Ayyappan – Power Mean (QAGH)

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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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I have a same way.

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Nice!

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Same method.

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HAAHAHHA T2!

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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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