Proof: Let
f(a,b,c)=a4+b4+c4+ab3+bc3+ca3−2(a3b+b3c+c3a)
Then
f(a+t,b+t,c+t)=6(∑a2−∑ab)t2+3(∑a3+∑cr a2b−2∑cycab2)t+a4+b4+c4+ab3+bc3+ca3−2(a3b+b3c+c3a)
Thus, it suffices to prove
Δ(a,b,c)=3(∑a3+∑qca2b−2∑qcab2)2−4⋅6(∑a2−∑ab)⋅(a4+b4+c4+ab3+bc3+ca3−2(a3b+b3c+c3a))⩽0
Notice that Δ(a,b,c)=Δ(a+t,b+t,c+t),t∈R, so it suffices to prove
Δ(a−c,b−c,0)⩽0⇔(3(a−c)2(b−c)−6(a−c)(b−c)2+3(a−c)3+3(b−c)3)2−4(−3(a−c)(b−c)+3(a−c)2+3(b−c)2)((a−c)4+(b−c)4+(a−c)3(b−c)−2(a−c)(b−c)3)⩽0⇔−3((b−c)3−3(a−c)2(b−c)+(a−c)3)2⩽0
The proposition is proved.