We have : \(x\le1\)\(\Rightarrow\left\{{}\begin{matrix}x^2\le x\\\sqrt{x}\le1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2\le x\\x\le\sqrt{x}\end{matrix}\right.\)\(\Rightarrow\sqrt{x}\ge x^2\)

\(\Rightarrow y\sqrt{x}+\dfrac{1}{4}\ge yx^2+\dfrac{1}{4}\) (the equality happens when x = 0 or x = 1)

Moreover, \(yx^2+\dfrac{1}{4}\ge2\sqrt{\dfrac{1}{4}yx^2}=x\sqrt{y}\)

(the equality happens only when \(yx^2=\dfrac{1}{4}\))

So, \(y\sqrt{x}+\dfrac{1}{4}\ge x\sqrt{y}\) or \(x\sqrt{y}-y\sqrt{x}\le\dfrac{1}{4}\)

The equality happens when : \(\left\{{}\begin{matrix}x=1\\yx^2=\dfrac{1}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{4}\end{matrix}\right.\)