Lời giải:
Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{x^2+xy}+\frac{1}{y^2+xy}\geq \frac{4}{x^2+xy+y^2+xy}=\frac{4}{(x+y)^2}\geq \frac{4}{1^2}=4\)

Ta có đpcm

Dấu "=" xảy ra khi $x=y=\frac{1}{2}$

Xét \(\frac{x}{y^3-1}+\frac{y}{x^3-1}=\frac{1-y}{y^3-1}+\frac{1-x}{x^3-1}=-\frac{1}{x^2+x+1}-\frac{1}{y^2+y+1}\)

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

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

từ đó ta có đpcm

\(VT=\dfrac{1}{x^2+xy}+\dfrac{1}{y^2+xy}\)

\(\ge\dfrac{4}{x^2+2xy+y^2}\)

\(=\dfrac{4}{\left(x+y\right)^2}>4\)

Cách khác.

Ta có: \(A=\dfrac{1}{x\left(x+y\right)}+\dfrac{1}{y\left(x+y\right)}=\dfrac{1}{x+y}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(=\dfrac{1}{x+y}.\dfrac{x+y}{xy}=\dfrac{1}{xy}\)

Áp dụng BĐT cho các số x,y >0 , ta có:

\(x+y\ge2\sqrt{xy}\Rightarrow\left(x+y\right)^2\ge4xy\Rightarrow\dfrac{\left(x+y\right)^2}{4}\ge xy\)

Và x+y \(\le\)1 \(\Rightarrow xy\le\dfrac{1}{4}\) \(\Rightarrow A\ge\dfrac{1}{\dfrac{1}{4}}=4\)

Dấu ''='' xảy ra khi x = y =0,5

\(B=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2+\left(xy+\frac{1}{xy}\right)^2\)

\(-\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\left(xy+\frac{1}{xy}\right)\)

\(\Rightarrow B=x^2+2+\frac{1}{x^2}+y^2+2+\frac{1}{y^2}+x^2y^2+2+\frac{1}{x^2y^2}-x^2y^2\) 

\(-2-x^2-y^2-\frac{1}{y^2}-\frac{1}{x^2}-\frac{1}{x^2y^2}\)

\(\Rightarrow B=x^2y^2-x^2y^2+x^2-x^2+1.\frac{1}{x^2}+1.\frac{1}{x^2y^2}-1.\frac{1}{x^2}-1\)

\(.\frac{1}{x^2y^2}+1.\frac{1}{y^2}-1.\frac{1}{y^2}+y^2-y^2+2+2+2-2\)

\(\Rightarrow B=4\)