Proof of Triangle Inequality: a+b-c, b+c-a, c+a-b

[Albert1]

Let a, b, c be the lengths of the sides of a triangle. Prove that:
$\sqrt{a+b-c}$+$\sqrt{b+c-a}$+$\sqrt{c+a-b}\leq\sqrt{a}+\sqrt{b}+\sqrt{c}$

 

[Albert1]

Let a, b, c be the lengths of the sides of a triangle. Prove that:
$\sqrt{a+b-c}$+$\sqrt{b+c-a}$+$\sqrt{c+a-b}\leq\sqrt{a}+\sqrt{b}+\sqrt{c}$

$(\sqrt{a+b-c}$+$\sqrt{b+c-a})^2\leq 4b$
$\therefore \sqrt{a+b-c}$+$\sqrt{b+c-a}\leq 2\sqrt{b}--------------(1)$
likewise
$\therefore \sqrt{b+c-a}$+$\sqrt{c+a-b}\leq 2\sqrt{c}--------------(2)$
$\therefore \sqrt{a+b-c}$+$\sqrt{c+a-b}\leq 2\sqrt{a}--------------(3)$
(1)+(2)+(3) the proof is done