RMO practice 2

Prove that,2a+b+2b+c+2c+a9a+b+ca,b,cR+\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a} \geq \dfrac{9}{a+b+c}\\a,b,c \in R^+

#Inequalities #RMO

Note by Adarsh Kumar
5 years, 7 months ago

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Comments

Using T2's Lemma,

2a+b+2b+c+2c+a=(2)2a+b+(2)2b+c+(2)2c+a(32)22(a+b+c)=182(a+b+c)=9a+b+c ¨\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 - 5 years, 7 months ago

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HAAHAHHA T2!

Pi Han Goh - 5 years, 7 months ago

Same method.

Swapnil Das - 5 years, 7 months ago

Nice!

Adarsh Kumar - 5 years, 7 months ago

I have a same way.

Dev Sharma - 5 years, 7 months ago

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

Pi Han Goh - 5 years, 7 months ago

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T2 is a cuter method though :P

Swapnil Das - 5 years, 7 months ago

From your inequality, we have: 2(a+b+c)(1a+b+1b+c+1c+a)92(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 Minh Quân - 5 years, 7 months ago

can someone tell me what is T2's Lemma?

Ganesh Ayyappan - 5 years, 7 months ago

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Titu's Lemma

Pi Han Goh - 5 years, 7 months ago

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

Ganesh Ayyappan - 5 years, 7 months ago

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@Ganesh Ayyappan Power Mean (QAGH)

Pi Han Goh - 5 years, 7 months ago
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