(1)
a+b=[a+b－2√(ab)]+2√(ab)=(√a－√b)²+2√(ab)
∵(√a－√b)²≥0
∴a+b≥2√(ab)
同理a+c≥2√(ac),b+c≥2√(bc)
∴(a+b)(b+c)(c+a)>=2√(ab)×2√(ac)×2√(bc)=8abc
(2)(a/b)+(b/c)+(c/a)>=3
令a/b=p^3,b/c=q^3,c/a=r^3
(a/b)+(b/c)+(c/a)=p^3＋q^3＋r^3
现证明p^3＋q^3＋r^3≥3pqr
p^3+q^3+r^3-3pqr
=[( p+q)^3-3p^2q-3pq^2]+r^3-3pqr
=[(p+q)^3+r^3]-(3p^2q+3pq^2+3pqr)
=(p+q+r)[(p+q)^2-(p+q)r+r^2]-3pq(p+q+r)
=(p+q+r)(p^2+q^2+2pq-pr-qr+r^2)-3ab(p+q+r)
=(p+q+r)(p^2+q^2+r^2-pq-pr-qr)
=(p+q+r)(2p^2+2q^2+2r^2-2pq-2qr-2pr)/2
=(p+q+r)[(p－q)^2+(q-r)^2＋（p－r）^2]/2
p+q+r都为正实数,
所以p^3+q^3+r^3-3pqr＝(p+q+r)[(p－q)^2+(q-r)^2＋（p－r）^2]/2≥0
当且仅当p＝q＝r时,p^3+q^3+r^3-3pqr＝0
即p^3＋q^3＋r^3≥3pqr成立
把a/b=p^3,b/c=q^3,c/a=r^3代入有
(a/b)+(b/c)+(c/a)≥3×a/b×b/c×c/a=3
即(a/b)+(b/c)+(c/a)≥3得证