Problem
ALG-B1-M01-P025 Strong Form of the Cubic Identity
#25
★★★★★ Level 5 of 5
Let \(a,b,c\ge0\). Prove that \(a^3+b^3+c^3\ge3abc\), and determine when equality holds.
Use the factorization of \(a^3+b^3+c^3-3abc\).
We have \[ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca). \] Since \(a,b,c\ge0\), the first factor \(a+b+c\ge0\).
The second factor is non-negative because \[ 2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2. \] Hence the whole difference is non-negative. Equality occurs when \(a+b+c=0\), which gives \(a=b=c=0\), or when the second factor is zero, which gives \(a=b=c\). Thus equality holds exactly when \(a=b=c\).
A good problem on turning an identity into an inequality and equality case.