MathLabs

Problem 5

Let a,b,ca,b,c be the side lengths of a triangle. Prove that a+b−c+b+c−a+c+a−b≤a+b+c\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\le\sqrt a+\sqrt b+\sqrt c, and determine when equality occurs.
Step 2 of 5: Apply the lemma to recover the term sqrt a
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a=(a+b−c)+(c+a−b)2≥a+b−c+c+a−b2\sqrt{a}=\sqrt{\frac{(a+b-c)+(c+a-b)}{2}}\ge\frac{\sqrt{a+b-c}+\sqrt{c+a-b}}{2}
Detailed analysis

The triangle inequalities make all three radicands positive. Since (a+b−c)+(c+a−b)=2a(a+b-c)+(c+a-b)=2a, the lemma with x=a+b−cx=a+b-c and y=c+a−by=c+a-b gives the displayed bound for a\sqrt a.