请教两道不等式证明题：1、若x,y,z属于R+,且x+y+z=xyz,证明不等式(y+z)/x+

1.不等式等价于xyz(xy(x+y)+yz(y+z)+zx(z+x)) ≥ 2(xy+yz+zx)².
由xyz = x+y+z,进一步等价于(x+y+z)(xy(x+y)+yz(y+z)+zx(z+x)) ≥ 2(xy+yz+zx)².
也即((x+y)+(y+z)+(z+x))(z²(x+y)+x²(y+z)+y²(z+x)) ≥ (z(x+y)+x(y+z)+y(z+x))².
易见这由Cauchy不等式立即得到.
2.由对称性,不妨设a ≤ b ≤ c.
则a/(bc+1)+b/(ca+1) ≤ a/(ab+1)+b/(ab+1) = (a+b)/(ab+1) = 1-(1-a)(1-b)/(ab+1) ≤ 1.
又c/(ab+1) ≤ c ≤ 1.
相加即得a/(bc+1)+b/(ca+1)+c/(ab+1) ≤ 2.