[tex]P=\frac{x^4+y^4}{x^4-y^4}-\frac{xy}{x^2-y^2}+\frac{x+y}{2.(x-y)}\\\\ =\frac{(x+y)^2-2xy}{2.(x-y)}+\frac{x^4+y^4}{x^4-y^4}\\\\ =\frac{x^2+y^2}{2.(x^2-y^2)}+\frac{x^4+y^4}{x^4-y^4}\\\\ =\frac{1}{2}.(\frac{x^2+y^2}{x^2-y^2})+\frac{x^4+y^4}{x^4-y^4}\\\\ =\frac{1}{2}.(\frac{x^2-y^2}{x^2-y^2}+\frac{2y^2}{x^2-y^2})+\frac{x^4-y^4}{x^4-y^4}+\frac{2y^4}{x^4-y^4}\\\\ =\frac{1}{2}.(1+\frac{2y^2}{x^2-y^2})+1+\frac{2y^4}{x^4-y^4}\\\\ \geq \frac{1}{2}.1+1=1,5[/tex]
dấu "=" xảy ra <=> y=0 và x bất kì >0