Lời giải:
BĐT cần chứng mình tương đương với:

$(xy+yz+xz)^2\geq 3(x+y+z)$

$\Leftrightarrow (xy+yz+xz)^2\geq 3xyz(x+y+z)$

$\Leftrightarrow (xy)^2+(yz)^2+(zx)^2+2xyz(x+y+z)\geq 3xyz(x+y+z)$

$\Leftrightarrow (xy)^2+(yz)^2+(xz)^2\geq xyz(x+y+z)$

$\Leftrightarrow (xy)^2+(yz)^2+(xz)^2-xyz(x+y+z)\geq 0$

$\Leftrightarrow 2(xy)^2+2(yz)^2+2(xz)^2-2xyz(x+y+z)\geq 0$

$\Leftrightarrow (xy-yz)^2+(yz-xz)^2+(xz-xy)^2\geq 0$

(luôn đúng với mọi $x,y,z\geq 0$)

Dấu "=" xảy ra khi $x=y=z=1$

\(\frac{2013x}{xy+2013x+2013}+\frac{y}{yz+y+2013}+\frac{z}{xz+z+1}\)

\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)

\(=\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}\)

\(=\frac{xz+z+1}{xz+z+1}=1\)

=>đpcm

2013x/xy+2013x+2013 + y/yz+y+2013 + z/xz+z+1

= xyz.x/xy+xyz.x+xyz + y/yz+y+xyz + z/xz+z+1

= xz/1+xz+z + 1/z+1+xz + z/xz+z+1

= xz+1+x/1+xz+x = 1 (đpcm)

Thay xyz=2013 vào ta có:

\(\frac{xyz\cdot x}{xy+xyz\cdot x+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)

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

\(=\frac{xy\cdot xz}{xy\left(xz+z+1\right)}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)

\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)

\(=\frac{xz+1+z}{xz+z+1}=1\) (Đpcm)

x²=yz  => \(\frac{x}{y}=\frac{z}{x}\)

\(z^2=xy\Rightarrow\frac{z}{x}=\frac{y}{z}\)

\(\Rightarrow\frac{x}{y}=\frac{z}{x}=\frac{y}{z}\)

áp dụng ... ta có

\(\frac{x}{y}=\frac{z}{x}=\frac{y}{z}=\frac{x+z+y}{y+x+z}=1\)

\(\frac{x}{y}=1\Rightarrow x=y\)

\(\frac{z}{x}=1\Rightarrow z=x\)

=>x=y=z

Ta có x2=yz nên x/y=z/x(1)

y2=xz nên x/y=y/z(2)

z2=xy nên z/x=y/z(3)

Từ 1,2,3 suy ra x/y=z/x=y/z(4)

áp dụng t/c dãy tỉ số bằng nhau vào 4 có

x/y=z/x=y/z=x+y+z/x+y+z

vì x, y,z khác 0 nên x+y+z Khác 0

suy ra x+y+z/z+x+y=1

suy ra x/y=z/x=y/z=1

suy ra x=y; x=z; y=z

xz=yx" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax">yx=zy" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax">zy=xz" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax">xz=yx=zy" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax">xz=yx=zy=x+y+zz+x+y=1" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax">xz=1⇒x=z" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax">yx=1⇒y=x" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax">zy=1⇒z=y" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax">zy=1⇒z=y" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline; float:none; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax">T

Vì 0xy+yz+xz=0.Nên:X,y,z đều bằng 0 và bằng nhau.

Đặt x² = yz (1) ; y² = xz (2) ; z² = xy (3)

Từ (1) => z= x²/y. Từ (2) => z = y²/x => x²/y = y²/x => x³ = y³ => x = y (*)

Tương tự : Từ (1) => y =x²/z. Từ (3) => y = z²/x => x²/z = z²/x => x³ = z³ => x = z(**)

Từ (*) và (**) suy ra x = y = z

thanks :)))

có thể làm được cách nhanh hơn thử làm theo tính chất dãy tỉ số bằng nhau hay hơn...