-Đề sai.

Giả sử \(x=\dfrac{1}{3};y=\dfrac{2}{3};z=1\Rightarrow x+y+z=2\)

\(\dfrac{1}{3}.\dfrac{2}{3}+2.\dfrac{2}{3}.1+2.1.\dfrac{1}{3}=\dfrac{20}{9}< 3\)

\(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{x-y}{-1};\dfrac{y}{2019}=\dfrac{z}{2020}=\dfrac{y-z}{-1};\dfrac{x}{2018}=\dfrac{z}{2020}=\dfrac{x-z}{-2}\\ \Leftrightarrow\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-2}\\ \Leftrightarrow2\left(x-y\right)=2\left(y-z\right)=x-z\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)

\(x+\sqrt{3x+yz}=x+\sqrt{x\left(x+y+z\right)+yz}=x+\sqrt{\left(x+y\right)\left(z+x\right)}\ge x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}\)

\(=x+\sqrt{xz}+\sqrt{xy}=\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Tương tự:

\(\dfrac{y}{y+\sqrt{3y+zx}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng vế với vế ta có đpcm

Cách này đòi hỏi sự kiên nhẫn và kinh nghiệm.

Cần chứng minh:

\({\dfrac {4 \left( xy+zx+yz \right) \left( x+y+z \right) ^{7}}{ 243}}- \left( {x}^{3}+{y}^{3}+{z}^{3} \right) \left( {x}^{3}{y}^{3}+{ x}^{3}{z}^{3}+{y}^{3}{z}^{3} \right) \geqslant 0.\quad(1) \) 

Đặt 

\(\text{M}=4\,{z}^{7}+ \left( 757\,x+757\,y \right) {z}^{6}+84\, \left( x+y \right) ^{2}{z}^{5}+140\, \left( x+y \right) ^{3}{z}^{4}\\\quad\quad+ \left( 1598 \,{x}^{4}+4205\,{x}^{3}y+4971\,{x}^{2}{y}^{2}+4205\,x{y}^{3}+1598\,{y} ^{4} \right) {z}^{3}\\\quad \quad+84\, \left( x+y \right) ^{5}{z}^{2}+28\, \left( x +y \right) ^{6}z\geqslant 0 \)

Ta có:

\((1)\Leftrightarrow \dfrac{1}{243}xy\cdot M+{\dfrac { \left( x+y \right) \left( {x}^{2}+11\,xy+{y}^{2} \right) \left( 2\,x-y \right) ^{2} \left( x-2\,y \right) ^{2}xy}{243}}\\\quad\quad+{ \dfrac { \left( x+y \right) z \left( x+y+z \right) \left( {x}^{2}+2\,x y+11\,zx+{y}^{2}+11\,yz+{z}^{2} \right) \left( 2\,y-z+2\,x \right) ^{ 2} \left( y-2\,z+x \right) ^{2}}{243}}\geqslant 0. \)

Đẳng thức xảy ra khi $...$

Ta co x+2/2=y+3/3=z+4/4  hay  x+1=y+1=z+1  =>  x=y=z

Suy ra:   2x+y+z=11 hay  2x+x+x=11  =>  4x=11  =>  x=11/4

Vay: x^2+y^2+z^2 = (11/4)^2+(11/4)^2+(11/4)^2 =121/16 . 3 = 363/16                                        

với mọi x, y, z ta có: 
(x-y)^2 +(y-z)^2+ (z-x)^2>=0 
<=>2x^2 +2y^2 + 2z^2 - 2xy -2yz - 2xz >=0 
<=>x^2 + y^2 +z^2 - xy -yz -zx >=0 
<=>(x+y+z)^2 >= 3(x+y+z) 
<=>[(x+y+z)^2]/3 >= xy+yz+ zx 
=>xy +yz + zx <=3 
dấu = xảy ra khi x=y=z =1

ai tích mình tích lại nhưng phải lên điểm mình tích gấp đôi

bạn kia có chỗ sai nha.