MathLabs

Problem 6

Let a,b,ca,b,c be the side lengths of a triangle. Prove that a2b(a−b)+b2c(b−c)+c2a(c−a)≥0a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge0. Determine when equality occurs.
Step 2 of 5: Simplify the cyclic expression
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a2b(a−b)+b2c(b−c)+c2a(c−a)=2(xy3+yz3+zx3−xyz(x+y+z))a^2b(a-b)+b^2c(b-c)+c^2a(c-a)=2\bigl(xy^3+yz^3+zx^3-xyz(x+y+z)\bigr)
Detailed analysis

Substituting the three Ravi expressions and expanding gives exactly the displayed identity. Therefore it remains to prove xy3+yz3+zx3≥xyz(x+y+z)xy^3+yz^3+zx^3\ge xyz(x+y+z).