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

AI Thread Summary
The discussion focuses on proving the inequality involving the sides of a triangle, specifically that the sum of the square roots of the expressions $\sqrt{a+b-c}$, $\sqrt{b+c-a}$, and $\sqrt{c+a-b}$ is less than or equal to the sum of the square roots of the sides themselves, $\sqrt{a}+\sqrt{b}+\sqrt{c}$. The proof involves manipulating the inequalities for pairs of terms, showing that each pair is bounded by twice the square root of the remaining side. By applying this reasoning to all combinations of the triangle's sides, the conclusion follows that the overall inequality holds true. The mathematical steps are clearly outlined, leading to the final proof of the inequality. This establishes a fundamental relationship between the sides of a triangle and their respective square roots.
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Let a, b, c be the lengths of the sides of a triangle. Prove that:
$\sqrt{a+b-c}$+$\sqrt{b+c-a}$+$\sqrt{c+a-b}\leq\sqrt{a}+\sqrt{b}+\sqrt{c}$
 
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Albert said:
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
 

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