51. By Hölder's inequality, we have
(y2+z2x+z2+x2y+x2+y2z)2[x(y2+z2)+y(z2+x2)+z(x2+y2)]⩾(x+y+z)3
It suffices to prove that
(x+y+z)3>4[x(y2+z2)+y(z2+x2)+z(x2+y2)]=4(x+y+z)(xy+yz+zx)−12xyz
By Schur's inequality, (x+y+z)3−4(x+y+z)(xy+yz+zx)+9xyz⩾0, so
(x+y+z)3⩾4(x+y+z)(xy+yz+zx)−9xyz>4(x+y+z)(xy+yz+zx)−12xyz