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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

$ \sqrt {ab}+\sqrt {bc}+\sqrt {ca}\leq a+b+c<2\sqrt {ab}+2\sqrt {bc}+2\sqrt {ca}$

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- Thread starter
- #1

- Jan 25, 2013

- 1,225

$ \sqrt {ab}+\sqrt {bc}+\sqrt {ca}\leq a+b+c<2\sqrt {ab}+2\sqrt {bc}+2\sqrt {ca}$

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- #2

- Jan 25, 2013

- 1,225

proof of left side:

$ \sqrt {ab}+\sqrt {bc}+\sqrt {ca}\leq a+b+c<2\sqrt {ab}+2\sqrt {bc}+2\sqrt {ca}$

$2\sqrt {ab}\leq a+b----(1)$

$2\sqrt {bc}\leq b+c----(2)$

$2\sqrt {ca}\leq c+a----(3)$

(1)+(2)+(3):$2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})\leq 2(a+b+c)$

$\therefore \sqrt {ab}+\sqrt {bc}+\sqrt {ca}\leq a+b+c$

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- Jan 25, 2013

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proof of right side:proof of left side:

$2\sqrt {ab}\leq a+b----(1)$

$2\sqrt {bc}\leq b+c----(2)$

$2\sqrt {ca}\leq c+a----(3)$

(1)+(2)+(3):$2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})\leq 2(a+b+c)$

$\therefore \sqrt {ab}+\sqrt {bc}+\sqrt {ca}\leq a+b+c$

let: $a\leq b\leq c$

$\sqrt {ab}+\sqrt {bc}\geq a+b-----(4)$

$\sqrt {bc}+\sqrt {ca}\geq a+b-----(5)$

$\sqrt {ca}+\sqrt {ab}\geq a+a-----(6)$

(4)+(5)+(6):

$2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})> 2a+2c>a+b+c$