ÁD BĐT AM-GM:$6(x^{2}+y^{2}+z^{2})+6xyz+30-18(x+y+z)=6(\Sigma x^{2})+3(2xyz+1)+27-3.2.3.(x+y+z)≥6(Σx2)+93√x2y2z2+27−3[(x+y+z)2+9]\geq 3(\Sigma x^{2})+\frac{27}{x+y+z}-6(xy+yz+zx)(*)TheoBĐTShur:\frac{9}{x+y+z}\geq4(xy+yz+zx)-(x+y+z)^{2}=2(xy+yz+zx)-(\Sigma x^{2})$$\Rightarrow(*) \geq0\Rightarrow đpcm$
ÁD BĐT AM-GM:$6(x^{2}+y^{2}+z^{2})+6xyz+30-18(x+y+z)=6(\Sigma x^{2})+3(2xyz+1)+27-3.3.3.(x+y+z)≥6(Σx2)+93√x2y2z2+27−3[(x+y+z)2+9]\geq 3(\Sigma x^{2})+\frac{27}{x+y+z}-6(xy+yz+zx)(*)TheoBĐTShur:\frac{9}{x+y+z}\geq4(xy+yz+zx)-(x+y+z)^{2}=2(xy+yz+zx)-(\Sigma x^{2})$$\Rightarrow(*) \geq0\Rightarrow đpcm$
ÁD BĐT AM-GM:$6(x^{2}+y^{2}+z^{2})+6xyz+30-18(x+y+z)=6(\Sigma x^{2})+3(2xyz+1)+27-3.
2.3.(x+y+z)
≥6(Σx2)+93√x2y2z2+27−3[(x+y+z)2+9]\geq 3(\Sigma x^{2})+\frac{27}{x+y+z}-6(xy+yz+zx)(*)
TheoBĐTShur:\frac{9}{x+y+z}\geq4(xy+yz+zx)-(x+y+z)^{2}=2(xy+yz+zx)-(\Sigma x^{2})$$\Rightarrow(*) \geq0\Rightarrow đpcm$