分类：教育/科学-升学入学-高考
问题：一道不等式巳知x,y,z为正数，且xyz=1。求证:(x+y+z? 
一道不等式
巳知x,y,z为正数，且xyz=1。求证:
(x+y+z)*(yz+zx+xy)+6>=5(yz+zx+xy)

回答：巳知x,y,z为正数，且xyz=1。求证:
(x+y+z)*(yz+zx+xy)+6>=5(yz+zx+xy)

证明 设a,b,c为正实数，令x=bc/a,y=ca/b,z=ab/c，对所证不等式作置换得:
(a^3+b^3+c^3)*(b^3*c^3+c^3*a^3+a^3*b^3)+6(abc)^3≥5abc(b^3*c^3+c^3*a^3+a^3*b^3)
<==> Σa^6*(b^3+c^3)-5abcΣb^3*c^3+9(abc)^3≥0
上式化简整理为
Σ[2b^3*c^3*(b+c)+2a^4*(b^3+b^2*c+bc^2+c^3)-3a^3*(bc+ca+ab)]*(b-c)^2≥0
<==>Σ[2b^3*c^3*(b+c)+2a^4*(b+c)*(b-c)^2+2a^4*bc(b+c)-3a^3*b^2*c^2]*(b-c)^2≥0
<==>Σbc[2b^2*c^2*(b+c)+2a^4*(b+c)-3a^3*bc]*(b-c)^2+2Σa^4*(b+c)*(b-c)^4≥0
据均值不等式易证
2b^2*c^2*(b+c)+2a^4*(b+c)≥8(b+c)*{[a^4/3]*[a^4/3]*[a^4/3]*[b^2*c^2]}^(1/4)
=8a^3*(b+c)√(bc)*(1/27)^(1/4)≥16a^3*bc*(1/27)^(1/4)>3a^3*bc.
故上式成立，证毕。