Đặt [tex]p=x+y+z,q=xy+yz+zx,r=xyz[/tex]
Ta thấy: [tex](x+y-z)(y+z-x)\leq (\frac{x+y-z+y+z-x}{2})^2=(\frac{2y}{2})^2=y^2[/tex]
Tương tự thì ta có:[tex](x+y-z)(x-y+z)\leq x^2,(y+z-x)(x+z-y)\leq z^2[/tex]
Nhân vế theo vế ta có:[tex](x+y-z)^2(y+z-x)^2(x+z-y)^2\leq x^2y^2z^2\Rightarrow (x+y-z)(y+z-x)(x+z-y)\leq xyz\Rightarrow xyz-(x+y-z)(y+z-x)(x+z-y)\geq 0\Rightarrow x^3 - x^2 y - x^2 z - x y^2 + 3 x y z - x z^2 + y^3 - y^2 z - y z^2 + z^3\geq 0\Rightarrow x^3+y^3+z^3+3xyz\geq xy(x+y)+yz(y+z)+zx(z+x)(1)[/tex]
Lại có:[tex](x+y)(y+z)(z+x)+xyz=(x+y+z)(xy+yz+zx)=pq\Rightarrow (x+y)(y+z)(z+x)=pq-r;xy(x+y)+yz(y+z)+zx(z+x)+3xyz=(x+y+z)(xy+yz+zx)=pq\Rightarrow xy(x+y)+yz(y+z)+zx(z+x)=pq-3r;x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(z+x)=p^3-3(pq-r)=p^3-3pq+3r[/tex]
BĐT (1) trở thành:[tex]p^3-3pq+3r+3r\geq pq-3r\Rightarrow p^3-4pq+9r\geq 0[/tex]
Mà [tex]p=x+y+z=1\Rightarrow 1-4q+9r\geq 0\Rightarrow 9r\geq 4q-1\Rightarrow r\geq \frac{4q-1}{9}\Rightarrow 2r\geq \frac{8q-2}{9}\Rightarrow xy+yz+zx-2xyz=q-2r\leq q-\frac{8q-2}{9}=\frac{q+2}{9}=\frac{xy+yz+zx+2}{9}\leq \frac{\frac{(x+y+z)^2}{3}+2}{9}=\frac{\frac{1}{3}+2}{9}=\frac{\frac{7}{3}}{9}=\frac{7}{27}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
