MathLabs

Problem 2

Suppose a,b,ca,b,c are the sides of a triangle. Prove that a2(b+c−a)+b2(c+a−b)+c2(a+b−c)≤3abca^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le3abc.
Step 4 of 4: Apply AM-GM
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x2y+x2z+y2x+y2z+z2x+z2y6≥xyz\frac{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}{6}\ge xyz
Detailed analysis

The six positive terms on the left have product (xyz)6(xyz)^6, so AM-GM gives their average at least xyzxyz. This proves the reduced inequality and hence the original claim.