\(VT=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\)

\(\ge\frac{3x}{y+z+1}+\frac{3y}{x+z+1}+\frac{3z}{x+y+1}\)

\(=\frac{3x^2}{xy+xz+x}+\frac{3y^2}{xy+yz+y}+\frac{3z^2}{xz+yz+z}\)

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

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

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

Dấu "=" <=> x=y=z=1

Theo đề ra ta có

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

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

Áp dụng tc của dãy tỉ số bằng nhau ta có

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

\(\Rightarrow\begin{cases}x=y\\y=z\\z=x\end{cases}\)

=> x=y=z (đpcm )

Ta có : \(x^2=yz;y^2=xz;z^2=xy\)

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

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

Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :

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

\(\Rightarrow x=y;y=z;z=x\)

\(\Rightarrow x=y=z\)

Cộng các đẳng thức trên với nhau được : 

\(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow2\left(x^2+y^2+z^2\right)=2\left(xy+yz+zx\right)\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\) 

Mà \(\left(x-y\right)^2\ge0\) , \(\left(y-z\right)^2\ge0\) , \(\left(z-x\right)^2\ge0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

Do đó dấu "=" xảy ra khi và chỉ khi x = y = z

Vậy x = y = z

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

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

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

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

=>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)

 Trước hết, ta đi chứng minh một bổ đề sau: Nếu \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\). Thật vậy, ta phân tích 

 \(P=a^3+b^3+c^3-3abc\)

\(P=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(P=\left(a+b+c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(P=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\).

Hiển nhiên nếu \(a+b+c=0\) thì \(P=0\) hay \(a^3+b^3+c^3=3abc\), bổ đề được chứng minh.

Do \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\) nên áp dụng bổ đề, ta được \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\).

Vì vậy \(\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}\) \(=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)\) \(=xyz.\dfrac{3}{xyz}=3\). Ta có 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

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