$x+y+z=xyz =>x^{2}+xy+xz=x^{2}yz=>yz(x^{2}+1)=(x+z)(x+y) A=\sqrt{\frac{4yz}{(x+y)(x+z)}}+\sqrt{\frac{2xz}{(x+y)2(y+z)}}+\sqrt{\frac{2xy}{(z+x)2(z+y)}}\leq \frac{y}{x+y}+\frac{z}{x+z}+\frac{x}{x+y}+\frac{z}{4(y+z)}+\frac{x}{x+z}+\frac{y}{4(y+z)}=9/4$
- Phuong Thu Quoc, leduylinh1998 và lahantaithe99 thích