áp dụng BĐT bunhiacopxki ,ta có :
\(\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)^2\le\left(x^2+y^2\right)\left(1-y^2+1-x^2\right)\)
↔\(1\le\left(x^2+y^2\right)\left(2-\left(x^2+y^2\right)\right)\)
đặt \(x^2+y^2=t\left(t\ge0\right)\),ta có:
\(t\left(2-t\right)\ge1\leftrightarrow t^2-2t+1\ge0\leftrightarrow\left(t-1\right)^2\ge0\)
dấu '=' xảy ra khi t=1 hay \(x^2+y^2=1\)