Vì x,y là số dương \(\Rightarrow\left\{{}\begin{matrix}y+0,5-y< y+0,5\\x+0,5-x< x+0,5\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2y}{y+0,5-y}>\dfrac{x^2y}{y+0,5}\\\dfrac{xy^2}{x+0,5-x}>\dfrac{xy^2}{x+0,5}\end{matrix}\right.\)\(\Rightarrow\dfrac{x^2y}{y+0,5}+\dfrac{xy^2}{x+0,5}< \dfrac{x^2y}{y+0,5-y}+\dfrac{xy^2}{x+0,5-x}=\dfrac{x^2y}{0,5}+\dfrac{xy^2}{0,5}=2x^2y+2xy^2=2xy\left(x+y\right)=2xy\cdot1=2xy\left(1\right)\)Đặt x=0,5+m; y=0,5+m thì x+y=0,5+m+0,5-m=1(thỏa mãn đề bài)

\(\Rightarrow xy=\left(0,5+m\right)\cdot\left(0,5-m\right)=0,5\cdot0,5+0,5m-0,5m-m\cdot m=0,25-m^2\)Vì:\(m^2\ge0\Rightarrow0,25-m^2\le0,25\Rightarrow xy\le0,25\Rightarrow2xy\le0,25\cdot2=0,5\left(2\right)\)Từ (1) và (2) \(\Rightarrow\dfrac{x^2y}{y+0,5}+\dfrac{xy^2}{x+0,5}< 0,5=\dfrac{1}{2}\)