\(x^2+y^2=x+y\\ \Leftrightarrow x^2-x+y^2-y=0\\ \Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{1}{2}\\ A=x+y=\left(x-\dfrac{1}{2}\right)+\left(y-\dfrac{1}{2}\right)+1\)
Áp dụng Bunhiacopski:
\(\left[\left(x-\dfrac{1}{2}\right)+\left(y-\dfrac{1}{2}\right)\right]^2\le\left(1^2+1^2\right)\left[\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2\right]=2\cdot\dfrac{1}{2}=1\\ \Leftrightarrow A\le1+1=2\)\(A_{max}=2\Leftrightarrow x=y=1\)
\(x^2+y^2\ge0\Rightarrow x+y=x^2+y^2\ge0\)
\(A_{min}=0\) khi \(x=y=0\)
Cách tìm max khác:
Ta có:
$(x-1)^2\geq 0, \forall x\in\mathbb{R}$
$\Rightarrow x^2+1\geq 2x$
Tương tự: $y^2+1\geq 2y$
$\Rightarrow 2(x+y)\leq x^2+y^2+2=x+y+2$
$\Rightarrow x+y\leq 2$ hay $A\leq 2$
Vậy $A_{\max}=2$ khi $x=y=1$
