# RMO practice 2

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

3 years, 5 months ago

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Using T2's Lemma,

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

- 3 years, 5 months ago

HAAHAHHA T2!

- 3 years, 5 months ago

Same method.

- 3 years, 5 months ago

Nice!

- 3 years, 5 months ago

I have a same way.

- 3 years, 5 months ago

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

- 3 years, 5 months ago

T2 is a cuter method though :P

- 3 years, 5 months ago

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 ^_^

- 3 years, 5 months ago

can someone tell me what is T2's Lemma?

- 3 years, 5 months ago

- 3 years, 5 months ago

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

- 3 years, 5 months ago

Power Mean (QAGH)

- 3 years, 5 months ago