不妨设
a1≥a2≥a3
则a1a2≥a1a3≥a2a3
(a1*a2)/a3+(a2*a3)/a1+(a3*a1)/a2≥a1+a2+a3
即证(a1a2)^2+(a2a3)^2+(a1a3^2)≥(a1+a2+a3)a1a2a3
(a1a2)^2+(a2a3)^2+(a1a3)^2≥a1^2a2a3+a1a2^2a3+a1a2a3^2
同序和≥乱序和
(a1a2)^2+(a2a3)^2+(a1a3)^2≥a1a2*a1a3+a1a2*a2a3+a1a3*a2a3
即(a1a2)^2+(a2a3)^2+(a1a3)^2≥a1^2a2a3+a1a2^2a3+a1a2a3^2
则原不等式得证