Question

x,y,z为正实数 x/(2x+y+z)+y/(x+2y+z）+z/(x+y+2z)<=3/4

Answer 1

x/(2x+y+z)=[3*(2x+y+z)-(x+2y+z）-(x+y+2z)-]/(4(2x+y+z))

y/(x+2y+z)=[3*(x+y+2z)-(2x+y+z)-(x+y+2z)]/(4(x+2y+z）)
z/(x+y+2z)=[3*(x+y+2z)-(2x+y+z)-(x+2y+z)]/(4(x+2y+z）)

所以
左边=（3/4）*3 -（1/4）（(2x+y+z)/(x+2y+z）+(x+2y+z）/(2x+y+z)+(x+2y+z）/(x+y+2z)+(x+y+2z)/(x+2y+z）+(x+y+2z)/(2x+y+z)+(2x+y+z)(x+y+2z)）
又烂段因为（(2x+y+z)/(x+2y+z）+(x+2y+z）/(2x+y+z)+(x+2y+z）/(x+y+2z)+(x+y+2z)/(x+2y+z）+(x+y+2z)/(2x+y+z)+(2x+y+z)(x+y+2z)）>=6

所以左边信历昌<=3/4

这个写的比较繁琐，你可以直接设(2x+y+z)=p，(x+2y+z）=q，(x+y+2z)=r
然后解出x，y，z，再把左边表滑扒示成p，q，r的形式，再利用基本不等式，是一样的。

Answer 2

左边=Σx^2/(2x^2+xy+xz)<=（柯西磨燃铅不等式的推论）(Σx)^2/(Σ(2x^2+xy+xz))=(Σx)^2/(2Σx^2+2Σxy)
由于(Σx)^2=Σx^2+2Σxy<=Σx^2+2Σx^2=3Σx^2
所以Σx^2>=(Σx)^2/3
所以左边段埋=(Σx)^2/(Σx^2+2Σxy+Σx^2)=(Σx)^2/((Σx)^2+Σx^2)<=(Σx)^2/((Σx)^2+(Σx)^2/瞎好3)=3/4