\(x^2+y^2< x+y\) . Dấu bé hơn hay dấu bé hơn hoặc bằng vậy bạn?
\(x^2+y^2\le x+y\)
\(\Rightarrow x^2-x+y^2-y\le0\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2\le\dfrac{1}{2}\left(1\right)\)
Theo BĐT Bunhiacopxki:
\(\left[\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2\right]\left(1+9\right)\ge\left[1.\left(x-\dfrac{1}{2}\right)+3.\left(y-\dfrac{1}{2}\right)\right]^2=\left(x+3y-2\right)^2\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{1}{2}.10\ge\left(x+3y-2\right)^2\)
\(\Rightarrow-\sqrt{5}\le x+3y-2\le\sqrt{5}\)
\(\Rightarrow-\sqrt{5}+2\le P\le\sqrt{5}+2\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{y-\dfrac{1}{2}}{3}\\\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}+\dfrac{1}{2\sqrt{5}}\\y=\dfrac{1}{2}+\dfrac{3}{2\sqrt{5}}\end{matrix}\right.\) (loại \(\left\{{}\begin{matrix}x=\dfrac{1}{2}-\dfrac{1}{2\sqrt{5}}\\y=\dfrac{1}{2}-\dfrac{3}{2\sqrt{5}}\end{matrix}\right.\) do \(y>0\) )
Và khi dấu bằng xảy ra thì \(P_{max}=\sqrt{5}+2\)
