\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)

\(VT\le\dfrac{x}{2\sqrt{x^2yz}}+\dfrac{y}{2\sqrt{y^2zx}}+\dfrac{z}{2\sqrt{z^2xy}}\)

\(VT\le\dfrac{1}{2}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\le\dfrac{1}{2}\sqrt{3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)}=\dfrac{\sqrt{3}}{2}\)

Dấu "=" xảy ra khi \(x=y=z=\sqrt{3}\)

Áp dụng BĐT AM-GM ta có:

\(\hept{\begin{cases}\sqrt{xy}\le\frac{x+y}{2}\\\sqrt{yz}\le\frac{y+z}{2}\\\sqrt{xz}\le\frac{x+z}{2}\end{cases}}\). Cộng theo vế ta có:

\(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=1\le\frac{x+y+y+z+x+z}{2}=\frac{2\left(x+y+z\right)}{2}=x+y+z\)

Do đó ta có: \(x+y+z\ge1\).Áp dụng BĐT Cauchy-Schwarz dạng Engel ta cũng có:

\(A\ge\frac{\left(x+y+z\right)^2}{x+y+y+z+x+z}=\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{1}{2}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)

13: 

xy(x+y)+yz(y+z)+xz(x+z)+2xyz 

= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz 

= xy(x + y) + yz(y + z + x) + xz(x + z + y) 

= xy(x + y) + z(x + y + z)(y + x) 

= (x + y)(xy + zx + zy + z²) 

= (x + y)[x(y + z) + z(y + z)] 

= (x + y)(y + z)(z + x)

x=3;y=5;z=6

xy = 15; yz = 30; zx = 18 

\(\Leftrightarrow\)xy . yz . zx = 15.30.18

\(\Leftrightarrow\)x² . y² . z² = 8100

\(\Leftrightarrow\)( x . y . z )² = 90²

\(\Leftrightarrow\) x . y . z  = 90

x = 90 : 15 = 6 hoặc x = -6

y = 90 : 30 = 3 hoặc y = -3

z = 90 : 18 = 5 hoặc z = -5

 vậy x = 6 hoặc x = -6

       y = 3 hoặc y = -3

       z = 5 hoặc z = -5