By the AM-GM inequality we have
i. a²b²+a²c² = a²(b²+c²) ≥ 2a²bc
ii. a²b²+b²c² = b²(a²+c²) ≥ 2b²ac
iii. a²c²+b²c² = c²(a²+b²) ≥ 2c²ab
Adding the above three inequalities gives
2a²b²+2a²c²+2b²c² ≥ 2a²bc+2b²ac+2c²ab
⇔a²b²+a²c²+b²c² ≥ abc(a+b+c)
as desired.