MHB Triangle Inequality Proof for Side Lengths of Triangle ABC

AI Thread Summary
The discussion focuses on proving the inequalities related to the side lengths of triangle ABC, specifically the expressions involving the square roots of the products of the sides. The left side of the inequality, \( \sqrt{ab} + \sqrt{bc} + \sqrt{ca} \leq a + b + c \), is established using the Cauchy-Schwarz inequality. For the right side, \( a + b + c < 2\sqrt{ab} + 2\sqrt{bc} + 2\sqrt{ca} \), the proof involves manipulating the triangle's properties and applying the AM-GM inequality. The discussion emphasizes the importance of these inequalities in understanding the relationships between the sides of a triangle. Overall, the proofs highlight fundamental geometric principles.
Albert1
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Triangle ABC with side lengths a,b,c please prove :

$ \sqrt {ab}+\sqrt {bc}+\sqrt {ca}\leq a+b+c<2\sqrt {ab}+2\sqrt {bc}+2\sqrt {ca}$
 
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Albert said:
Triangle ABC with side lengths a,b,c please prove :

$ \sqrt {ab}+\sqrt {bc}+\sqrt {ca}\leq a+b+c<2\sqrt {ab}+2\sqrt {bc}+2\sqrt {ca}$
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):$2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})\leq 2(a+b+c)$
$\therefore \sqrt {ab}+\sqrt {bc}+\sqrt {ca}\leq a+b+c$
 
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Albert said:
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):$2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})\leq 2(a+b+c)$
$\therefore \sqrt {ab}+\sqrt {bc}+\sqrt {ca}\leq a+b+c$
proof of right side:
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):
$2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})> 2a+2c>a+b+c$
 
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