Có \(VT=\dfrac{x^2}{x^3-xyz+2013x}+\dfrac{y^2}{y^3-xyz+2013y}+\dfrac{z^2}{z^3-xyz+2013z}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)

\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[x^2+y^2+z^2-\left(xy+yz+zx\right)\right]+2013\left(x+y+z\right)}\)

\(=\dfrac{x+y+z}{x^2+y^2+z^2-\left(xy+yz+zx\right)+3\left(xy+yz+zx\right)}\) 

(vì \(2013=3.671=3\left(xy+yz+zx\right)\))

\(=\dfrac{x+y+z}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}\)

\(=\dfrac{x+y+z}{\left(x+y+z\right)^2}\)

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

ĐTXR \(\Leftrightarrow\dfrac{1}{x^2-yz+2013}=\dfrac{1}{y^2-zx+2013}=\dfrac{1}{z^2-xy+2013}\)

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

\(\Leftrightarrow x=y=z\) (với \(x,y,z>0\))

Vậy ta có đpcm.

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

=> \(2x^2+2y^2+2x^2=2xy+2yz+2zx\) 

=> \(2x^2+2y^2+2x^2-2xy-2yz-2zx=0\) 

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

=> x -y =0 ; y - z=0 ; z - x=0

=> x =y; y =z; z=x

=> x=y=z

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=4\left(x^2+y^2+z^2-xy-xz-yz\right)\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2zy+z^2\right)+\left(z^2-2xz+x^2\right)=4\left(x^2+y^2+z^2-xy-xz-yz\right)\)

\(\Leftrightarrow2x^2-2xy+2y^2-2yz+2z^2-2xz=4\left(x^2+y^2+z^2-xy-yz-xz\right)\)

\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=4\left(x^2+y^2-xy-xz-yz\right)\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}}\)

\(\Leftrightarrow x=y=z\)

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=4.\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(< =>\left(x^2-2xy+y^2\right)+\left(y^2-2zy+z^2\right)+\left(z^2-2xz+x^2\right)=4.\left(x^2+y^2+z^2-xy-xz-yz\right)\)

\(< =>2x^2-2xy+2y^2-2yz+2z^2-2xz=4.\left(x^2+y^2+z^2-xy-xz-yz\right)\)

\(< =>2.\left(x^2+y^2+x^2-xy-xz-zy\right)=4.\left(x^2+y^2+z^2-xy-xz-zy\right)\)

\(< =>2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)

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

\(< =>\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\)

\(< =>\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}< =>x=y=z}\)

\(\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)

\(\Rightarrow2\left(x+y+z\right)\ge3\left(xy+yz+xz\right)\)

\(\Rightarrow x+y+z\ge\dfrac{3}{2}\left(xy+yz+xz\right)>xy+yz+xz\)(x,y,z>0)

Ta có:\(\sqrt{\dfrac{yz}{x^2+2017}}=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\)

  \(=\sqrt{\dfrac{y}{x+y}\cdot\dfrac{z}{x+z}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\)

Tương tự ta có:\(\sqrt{\dfrac{zx}{y^2+2017}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{z}{y+z}}{2}\)

                         \(\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)

Cộng vế với vế ta có:

\(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\)

\(\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}+\dfrac{z}{z+y}+\dfrac{x}{x+y}+\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)

\(=\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}}{2}=\dfrac{1+1+1}{2}=\dfrac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{\sqrt{2017}}{\sqrt{3}}\)

Áp dụng BĐT AM-GM ta có:

\(\frac{\left(y+z\right)\sqrt{yz}}{x}\ge\frac{2\sqrt{yz}\cdot\sqrt{yz}}{x}=\frac{2\sqrt{\left(yz\right)^2}}{x}=\frac{2yz}{x}\)

Tương tự cho 2 BĐT còn lại ta cũng có

\(\frac{\left(x+y\right)\sqrt{xy}}{z}\ge\frac{2xy}{z};\frac{\left(x+z\right)\sqrt{xz}}{y}\ge\frac{2xz}{y}\)

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

Cần chứng minh \(\frac{2xy}{z}+\frac{2yz}{x}+\frac{2xz}{y}\ge2\left(x+y+z\right)\)

\(\Leftrightarrow\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\ge x+y+z\)

Áp dụng BĐT AM-GM:

\(\frac{xy}{z}+\frac{yz}{x}\ge2\sqrt{\frac{xy}{z}\cdot\frac{yz}{x}}=2\sqrt{y^2}=2y\)

Tương tự rồi cộng theo vế ta có ĐPCM

Khi \(x=y=z\)

\(\dfrac{x-y}{z^2+1}=\dfrac{x-y}{z^2+xy+yz+zx}=\dfrac{x-y}{z\left(z+y\right)+x\left(z+y\right)}=\dfrac{x-y}{\left(x+z\right)\left(z+y\right)}\)

Tương tự: \(\dfrac{y-z}{x^2+1}=\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}\);\(\dfrac{z-x}{y^2+1}=\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)

Cộng vế với vế \(\Rightarrow VT=\dfrac{x-y}{\left(x+z\right)\left(y+z\right)}+\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}+\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)

\(=\dfrac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(=\dfrac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(đpcm)