Xét  \(A=x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)-3xyz\)

\(=\left(x+y+z\right)^3\left(x^2+y^2+z^2+2xy-xz-yz-3xy\right)\)

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

\(\Rightarrow2A=2\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

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

\(=\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]\)

Vì \(x+y+z\ge0\) ; \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\) với mọi  \(x,y,z\)

\(\Rightarrow2A\ge0\)

\(\Rightarrow A\ge0\)

\(\Rightarrow x^3+y^3+z^3\ge3xyz\)

Vậy nếu \(x+y+z\ge0\) thì \(x^3+y^3+z^3\ge3xyz\)

Lời giải:
Áp dụng BĐT AM-GM ta có:

\(\text{VT}=x-\frac{x}{x^2+z}+y-\frac{y}{y^2+x}+z-\frac{z}{z^2+y}=(x+y+z)-\left(\frac{x}{x^2+z}+\frac{y}{y^2+x}+\frac{z}{z^2+y}\right)\)

\(\geq (x+y+z)-\left(\frac{x}{2\sqrt{x^2z}}+\frac{y}{2\sqrt{y^2x}}+\frac{z}{2\sqrt{z^2y}}\right)=(x+y+z)-\frac{1}{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)(1)\)

Từ giả thiết \(xy+yz+xz=3xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)

Cauchy-Schwarz:

\(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3(2)\)

\(\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2\leq (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(1+1+1)=9\)

\(\Rightarrow \left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)\leq 3(3)\)

Từ \((1);(2);(3)\Rightarrow \text{VT}\geq 3-\frac{1}{2}.3=\frac{3}{2}\)

Mặt khác: \(\text{VP}=\frac{1}{2}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{2}\)

Do đó \(\text{VT}\geq \text{VP}\) (đpcm)

Dấu "=" xảy ra khi $x=y=z=1$

\(\Rightarrow\left(x+y+z\right)^2\ge\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\ge3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=\dfrac{3\left(x+y+z\right)}{xyz}\Rightarrow x+y+z\ge\dfrac{3}{xyz}\)

\(x+y+z=\dfrac{x+y+z}{3}+\dfrac{2\left(x+y+z\right)}{3}\ge\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{2}{3}.\dfrac{3}{xyz}\ge\dfrac{1}{3}\left(\dfrac{9}{x+y+z}\right)+\dfrac{2}{xyz}=\dfrac{3}{x+y+z}+\dfrac{2}{xyz}\left(đpcm\right)\)

\(dấu"="xảy\) \(ra\Leftrightarrow x=y=z=1\)

\(x^3+y^3-z^3\ge3xyz\)

\(\Leftrightarrow x^3+y^3+z^3-3xyz\ge0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\ge0\)

\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\ge0\)

Luôn đúng

Theo GT : \(xy+yz+xz=3xyz\Rightarrow\frac{xy+yz+xz}{xyz}=3\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)

\(\frac{x^3}{x^2+z}=\frac{x\left(x^2+z\right)}{x^2+z}-\frac{xz}{x^2+z}=x-\frac{xz}{x^2+z}\ge x-\frac{xz}{2x\sqrt{z}}=x-\frac{\sqrt{z}}{2}\)

Tương tự , ta có : \(\frac{y^3}{y^2+x}\ge y-\frac{\sqrt{x}}{2}\) ; \(\frac{z^3}{z^2+y}\ge z-\frac{\sqrt{y}}{2}\)

\(\Rightarrow\frac{x^3}{x^2+z}+\frac{y^3}{y^2+z}+\frac{z^3}{z^2+y}\ge x+y+z-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}\)

Vì x ; y ; z dương , áp dụng BĐT Cô - si , ta có :

\(x+1\ge2\sqrt{x};y+1\ge2\sqrt{y};z+1\ge2\sqrt{z}\)

\(\Rightarrow x+y+z+3\ge2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

=> \(\frac{x+y+z+3}{2}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}\) => BĐT được c/m

Tiếp tục AD BĐT Cô - si , ta có :

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{xyz}}=9\)

\(\Rightarrow x+y+z\ge\frac{9}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\frac{9}{3}=3\) => BĐT được c/m

Có : \(\frac{x^3}{x^2+z}+\frac{y^3}{y^2+x}+\frac{z^3}{z^2+y}\ge x+y+z-\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{2}\ge x+y+z-\frac{x+y+z+3}{4}=\frac{3x+3y+3z-3}{2}\ge\frac{3.3-3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=z=1\)

Vậy ...

\(a,\left(x+y+z\right)^3-x^3-y^3-z^3\\ =\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\\ =\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\\ =x^3+y^3+z^3+3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\\ =\left(x+y\right)\left(3xy+3xz+3yz+3z^2\right)\\ =3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\\ =3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

\(b,x^3+y^3+z^3-3xyz\\ =\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz+2xy-3xy\right)\\ =0\left(x^2+y^2+z^2-xz-yz-xy\right)=0\\ \Leftrightarrow x^3+y^3+z^3=3xyz\)