[不等式] 一个不等式

实数$a,b,c>0$, 证明$a^5+b^5+c^5+abc(ab+bc+ca)\ge a^3b^2+a^3c^2+b^3c^2+b^3a^2+c^3a^2+c^3b^2$.

1# 力工
proof:
\[ \Leftrightarrow a(a^2-b^2)(a^2-c^2)+b(b^2-a^2)(b^2-c^2)+c(c^2-a^2)(c^2-b^2)\geq 0 \]
Which is perfectly true due to Schur inequality.