>>628
(1/6)(p-p')^2･(ab+bc+ca)^2 ≧0 を足すと…

〔系〕
a^4 +b^4 +c^4 - p(aaab+bbbc+ccca) - p'(abbb+bccc+caaa) + {(pp+p'p')/2 -1}(aabb+bbcc+ccaa) + (p+p'-pp')abc(a+b+c)≧0,