\(xy\ne0,x,y\ne1\)

\(A=\dfrac{x^{ }}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2\left(x+y\right)}{x^2y^2+3}\)

\(xét:\dfrac{2\left(x+y\right)}{x^2y^2+3}=\dfrac{2}{x^2y^2+3}\left(1\right)\)

\(\dfrac{x^{ }}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{x^4-x-y^4+y}{\left(x^3-1\right)\left(y^3-1\right)}\left(2\right)\)

\(xét:\) \(x^4-x-y^4+y=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3-1\right)\)

\(=\left(x-y\right)\left[\left(x+y\right)^3-3xy\left(x+y\right)+xy\left(x+y\right)-1\right]\)

\(=\left(x-y\right)\left(1-3xy+xy-1\right)\)

\(=\left(x-y\right)\left(-2xy\right)=-2xy\left(x-y\right)=2xy\)

\(xét\) \(\left(y^3-1\right)\left(x^3-1\right)=x^3y^3-\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+1\)

\(=x^3y^3-\left(1-3xy\right)+1=x^3y^3+3xy=xy\left(x^2y^2+3\right)\)

\(\Rightarrow\left(2\right)\Leftrightarrow\dfrac{-2\left(x-y\right)}{x^2y^2+3}\)

\(\left(1\right)\left(2\right)\Rightarrow A=\dfrac{2}{x^2y^2+3}-\dfrac{2\left(x-y\right)}{x^2y^2+3}=\dfrac{2-2x+2y}{x^2y^2+3}\ne0\left(đề-sai\right)\)

Đặt \(\left(x;y;z\right)=\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\Rightarrow abc=1\)

\(P=\dfrac{a^2bc}{b+c}+\dfrac{ab^2c}{c+a}+\dfrac{abc^2}{a+b}=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(P=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\)

Ta có:

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

Do đó ta chỉ cần chứng minh:

\(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\ge\dfrac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)

Ta có:

\(\dfrac{x}{y}+\dfrac{x}{y}+1\ge3\sqrt[3]{\dfrac{x^2}{y^2}}\) 

Tương tự ...

Cộng lại ta có:

\(2\left(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\right)+6\ge3\left(\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\right)\)

\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\ge\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\)

Do đó ta chỉ cần chứng minh:

\(\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\ge\dfrac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)

\(\Leftrightarrow\left(\sqrt[3]{\dfrac{x}{y}}-\sqrt[3]{\dfrac{x}{z}}\right)^2+\left(\sqrt[3]{\dfrac{y}{x}}-\sqrt[3]{\dfrac{y}{z}}\right)^2+\left(\sqrt[3]{\dfrac{z}{x}}-\sqrt[3]{\dfrac{z}{y}}\right)^2\ge0\) (luôn đúng)

\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)

\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)

\(x=0;y=0\Leftrightarrow B=0\)

Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)

Vậy \(A\ne B\)