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 3 of 5: Apply the lemma to the other two sides
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b=(a+b−c)+(b+c−a)2≥a+b−c+b+c−a2,c=(b+c−a)+(c+a−b)2≥b+c−a+c+a−b2\sqrt{b}=\sqrt{\frac{(a+b-c)+(b+c-a)}{2}}\ge\frac{\sqrt{a+b-c}+\sqrt{b+c-a}}{2},\quad\sqrt{c}=\sqrt{\frac{(b+c-a)+(c+a-b)}{2}}\ge\frac{\sqrt{b+c-a}+\sqrt{c+a-b}}{2}
Detailed analysis

The same lemma with the other two pairs gives the displayed estimates for b\sqrt b and c\sqrt c. Add this step to the preceding one: each radicand square root appears exactly twice on the right and is divided by two.