Problem
ALG-B2-M10-P014 Fourth powers versus mixed terms
#14
★★★★★ Level 5 of 5
Prove for \(a,b,c\ge0\): \[a^4+b^4+c^4\ge abc(a+b+c).\]
Hint. Estimate \(a^2bc\), \(ab^2c\), and \(abc^2\) separately using four fourth powers.
By AM-GM, \[2a^4+b^4+c^4\ge4\sqrt[4]{a^8b^4c^4}=4a^2bc.\] Similarly, \[2b^4+c^4+a^4\ge4ab^2c,\qquad 2c^4+a^4+b^4\ge4abc^2.\] Adding the three estimates gives \[4(a^4+b^4+c^4)\ge4abc(a+b+c),\] which proves the claim.
Teaching goal: choose a combination of methods and always check the equality case.