Với $y < 0 < x$, so sánh $A = 2\left( {x - y} \right)x{y^3}.\dfrac{{\sqrt {{x^2}{y^3}} }}{{\sqrt {{x^4}{y^5}{{\left( {x - y} \right)}^2}} }}$ và $0.$

Với $y < 0 < x \Rightarrow x - y > 0$

Ta có: $\left| x \right| = x;\,\,\left| y \right| =  - y;\,\,\,\left| {x - y} \right| = x - y.$

$\begin{array}{l}A = 2\left( {x - y} \right)x{y^3}.\dfrac{{\sqrt {{x^2}{y^3}} }}{{\sqrt {{x^4}{y^5}{{\left( {x - y} \right)}^2}} }}\\ = 2\left( {x - y} \right)x{y^3}.\sqrt {\dfrac{{{x^2}{y^3}}}{{{x^4}{y^5}{{\left( {x - y} \right)}^2}}}} \\ = 2\left( {x - y} \right)x{y^3}.\sqrt {\dfrac{1}{{{x^2}{y^2}{{\left( {x - y} \right)}^2}}}} \\ = 2\left( {x - y} \right)x{y^3}.\dfrac{1}{{\sqrt {{x^2}{y^2}{{\left( {x - y} \right)}^2}} }}\\ = 2\left( {x - y} \right)x{y^3}.\dfrac{1}{{\left| x \right|.\left| y \right|.\left| {x - y} \right|}}\,\,\\ = \dfrac{{2\left( {x - y} \right)x{y^3}}}{{ - xy\left( {x - y} \right)}} =  - 2{y^2}\end{array}$

Vì ${y^2} > 0$ với mọi $y\ne 0$ $\Rightarrow  - 2{y^2} < 0$ với mọi $y\ne 0$

Vậy $A < 0.$