Triangle Inequality Proof for Side Lengths of Triangle ABC

[Albert1]

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}$

 

[Albert1]

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:

Spoiler

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

 

[Albert1]

proof of left side:

Spoiler

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

Spoiler

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