Problem
ALG-B2-M10-P004 Fourth-degree Schur
#4
★★★★★ Level 5 of 5
Prove for \(a,b,c\ge0\): \[\sum a^4+abc(a+b+c)\ge\sum_{\mathrm{sym}}a^3b.\]
Hint. This is Schur degree \(4\): check \(b=c\) and the boundary.
The inequality is symmetric. By UVW, it is enough to check \(b=c\) and the boundary. For \(b=c=1\), \(a=t\), the difference is \(t^2(t-1)^2\ge0\). For \(c=0\), we get \(a^4+b^4-a^3b-ab^3=(a-b)^2(a^2+ab+b^2)\ge0\). Hence the inequality is proved.
Teaching goal: choose a combination of methods and always check the equality case.