$P=\frac{1}{2+\frac{3y}{x}}+\frac{1}{1+\frac{z}{y}}+\frac{1}{1+\frac{x}{z}}$Do$x\geq y\Rightarrow \frac{x}{y}\geq1$ÁD BĐT$\frac{1}{1+\frac{z}{y}}+\frac{1}{1+\frac{x}{z}}\geq \frac{2}{1+\sqrt{\frac{x}{y}}}$ $\Rightarrow P\geq \frac{1}{2+\frac{3y}{x}}+\frac{2}{1+\sqrt{\frac{x}{y}}}$Đặt$t=\sqrt{\frac{x}{y}}(t\epsilon \left[ {1;2} \right])$$\Rightarrow P\geq \frac{t^{2}}{2t^{2}+3}+\frac{2}{1+t}=f(t)$$f'(t)=\frac{6t}{(2t^{2}+3)^{2}}-\frac{2}{(t+1)^{2}}\leq0,t\epsilon \left[ {1;2} \right]$$\Rightarrow min f(t)=f(2)=\frac{34}{33}$Dấu''='' xra$\Leftrightarrow x=4;y=1;z=2$

$P=\frac{1}{2+\frac{3y}{x}}+\frac{1}{1+\frac{z}{y}}+\frac{1}{1+\frac{x}{z}}$Do$x\geq y\Rightarrow \frac{x}{y}\geq1$ÁD BĐT$\frac{1}{1+\frac{z}{y}}+\frac{1}{1+\frac{x}{z}}\geq \frac{2}{1+\sqrt{\frac{x}{y}}}$ Đặt$t=\sqrt{\frac{x}{y}}(t\epsilon \left[ {1;2} \right])$$\Rightarrow P\geq \frac{t^{2}}{2t^{2}+3}+\frac{2}{1+t}=f(t)$$f'(t)= \frac{6t}{(2t^{2}+3)^{2}}-\frac{2}{(t+1)^{2}}\leq0,t\epsilon \left[ {1;2} \right]$$\Rightarrow min f(t)=f(2)=\frac{34}{33}$Dấu''='' xra$\Leftrightarrow x=4;y=1;z=2$

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)$$\geq6(\Sigma x^{2})+9\sqrt[3]{x^{2}y^{2}z^{2}}+27-3\left[ {(x+y+z)^{2}+9} \right]$$\geq 3(\Sigma x^{2})+\frac{27}{x+y+z}-6(xy+yz+zx)(*)$Theo BĐT Shur:$\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)$$\geq6(\Sigma x^{2})+9\sqrt[3]{x^{2}y^{2}z^{2}}+27-3\left[ {(x+y+z)^{2}+9} \right]$$\geq 3(\Sigma x^{2})+\frac{27}{x+y+z}-6(xy+yz+zx)(*)$Theo BĐT Shur:$\frac{9}{x+y+z}\geq4(xy+yz+zx)-(x+y+z)^{2}=2(xy+yz+zx)-(\Sigma x^{2})$$\Rightarrow(*) \geq0\Rightarrow đpcm$

$pt(2)\Leftrightarrow (a-b)(a^3-b^3-3a^2-3b^2)=2(a^3-b^3-3a^2-3b^2)$$\Leftrightarrow a^3-b^3=3(a^2+b^2)\Leftrightarrow (a-b)(a^2+ab+b^2)=3(a^2+b^2)$Mặt khác do $a-b<2 \vee ab \le |ab| \le a^2+b^2$\hookrightarrow VT <2(a^2+ab+b^2) \le 3(a^2+b^2)=VP$HPT VN

$pt(2)\Leftrightarrow (a-b)(a^3-b^3-3a^2-3b^2)=2(a^3-b^3-3a^2-3b^2)$$\Leftrightarrow a^3-b^3=3(a^2+b^2)\Leftrightarrow (a-b)(a^2+ab+b^2)=3(a^2+b^2)$Mặt khác do $a-b<2 \vee ab \le |ab| \le a^2+b^2$$\hookrightarrow VT <2(a^2+ab+b^2) \le 3(a^2+b^2)=VP$HPT VN