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

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

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TopNewestUsing T2's Lemma,

\[\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a} = \dfrac{\left(\sqrt{2}\right)^2}{a+b}+\dfrac{\left(\sqrt{2}\right)^2}{b+c}+\dfrac{\left(\sqrt{2}\right)^2}{c+a} \geq \dfrac{\left(3\sqrt{2}\right)^2}{2(a+b+c)} = \dfrac{18}{2(a+b+c)}= \dfrac{9}{a+b+c} \ \Huge{\ddot\smile}\] – Nihar Mahajan · 1 year ago

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– Dev Sharma · 1 year ago

I have a same way.Log in to reply

– Adarsh Kumar · 1 year ago

Nice!Log in to reply

– Swapnil Das · 1 year ago

Same method.Log in to reply

– Pi Han Goh · 1 year ago

HAAHAHHA T2!Log in to reply

Do AM-HM on the numbers \(a+b,b+c,c+a\). – Pi Han Goh · 1 year ago

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– Swapnil Das · 1 year ago

T2 is a cuter method though :PLog in to reply

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 ^_^ – Tôn Ngọc Quân · 1 year ago

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can someone tell me what is T2's Lemma? – Ganesh Ayyappan · 1 year ago

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Titu's Lemma – Pi Han Goh · 1 year ago

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– Ganesh Ayyappan · 1 year ago

Thank you ... how do u do by AM-HM inequality?Log in to reply

Power Mean (QAGH) – Pi Han Goh · 1 year ago

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