Theo bài ra: \(x+y=1\)
Ta có: \(\left(x-y\right)^2\ge0,\forall x;y\in R\)
\(\Rightarrow x\left(x-y\right)-y\left(x-y\right)\ge0,\forall x;y\in R\)
\(\Rightarrow x^2-xy-xy+y^2\ge0,\forall x;y\in R\)
\(\Rightarrow x^2+y^2-2xy\ge0,\forall x;y\in R\)
\(\Rightarrow x^2+y^2\ge2xy,\forall x;y\in R\)
\(\Rightarrow x^2+y^2+2xy\ge2xy+2xy,\forall x;y\in R\)
\(\Rightarrow\left(x^2+xy\right)+\left(y^2+xy\right)\ge4xy,\forall x;y\in R\)
\(\Rightarrow x\left(x+y\right)+y\left(x+y\right)\ge4xy,\forall x;y\in R\)
\(\Rightarrow\left(x+y\right)\left(x+y\right)\ge4xy,\forall x;y\in R\)
\(\Rightarrow\left(x+y\right)^2\ge4xy,\forall x;y\in R\)
\(\Rightarrow4xy\le1,\forall x;y\in R\)
\(\Rightarrow xy\le\dfrac{1}{4},\forall x;y\in R\)
\(\Rightarrow Max_{xy}=\dfrac{1}{4}\) khi \(\left\{{}\begin{matrix}x=y\\x+y=1\end{matrix}\right.\)
\(\Rightarrow Max_{xy}=\dfrac{1}{4}\) khi \(x=y=\dfrac{1}{2}\)
Vậy \(Max_{xy}=\dfrac{1}{4}\) khi \(x=y=\dfrac{1}{2}\)
