Let a, b, c>0, a+b+c=15.
The least value of E=a3a2+ab+b2+b3b2+bc+c2+c3c2+ca+a2 is
Observe that
a3−b3a2+ab+b2+b3−c3b2+bc+c2+c3−a3c2+ca+a2=(a−b)+(b−c)+(c−a)=0Thus, we can writeE=12(a3+b3a2+ab+b2+b3+c3b2+bc+c2+c3+a3c2+ca+a2)But a3+b3a2+ab+b2≥a+b3Since a2−2ab+b2=(a−b)2≥0∴E≥13(a+b+c)=5