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RMO practice 2

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
10 months, 2 weeks ago

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Using T2's Lemma,

\[\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a} = \dfrac{\left(\sqrt{2}\right)^2}{a+b}+\dfrac{\left(\sqrt{2}\right)^2}{b+c}+\dfrac{\left(\sqrt{2}\right)^2}{c+a} \geq \dfrac{\left(3\sqrt{2}\right)^2}{2(a+b+c)} = \dfrac{18}{2(a+b+c)}= \dfrac{9}{a+b+c} \ \Huge{\ddot\smile}\] Nihar Mahajan · 10 months, 2 weeks ago

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@Nihar Mahajan I have a same way. Dev Sharma · 10 months, 2 weeks ago

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@Nihar Mahajan Nice! Adarsh Kumar · 10 months, 2 weeks ago

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@Nihar Mahajan Same method. Swapnil Das · 10 months, 2 weeks ago

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@Nihar Mahajan HAAHAHHA T2! Pi Han Goh · 10 months, 2 weeks ago

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Do AM-HM on the numbers \(a+b,b+c,c+a\). Pi Han Goh · 10 months, 2 weeks ago

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@Pi Han Goh T2 is a cuter method though :P Swapnil Das · 10 months, 2 weeks ago

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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 ^_^ Tôn Ngọc Quân · 10 months, 1 week ago

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can someone tell me what is T2's Lemma? Ganesh Ayyappan · 10 months, 2 weeks ago

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@Ganesh Ayyappan Titu's Lemma Pi Han Goh · 10 months, 2 weeks ago

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@Pi Han Goh Thank you ... how do u do by AM-HM inequality? Ganesh Ayyappan · 10 months, 2 weeks ago

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@Ganesh Ayyappan Power Mean (QAGH) Pi Han Goh · 10 months, 1 week ago

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