Bài 2:
Ta có: ĐK: $x, y\geq 0$
$x\sqrt{1-y^2}+y\sqrt{1-x^2}=x^2+y^2<=>x(\sqrt{1-y^2}-x)+y(\sqrt{1-x^2}-y)=0<=>x\frac{1-y^2-x^2}{\sqrt{1-y^2}+x}+y\frac{1-y^2-x^2}{\sqrt{1-x^2}+y}=0<=>(1-x^2-y^2)(\frac{x}{\sqrt{1-y^2}+x}+\frac{y}{\sqrt{1-x^2}+y})=0$
mà $x, y\geq 0$ $<=>x^2+y^2=1$
Áp dụng BĐT Cauchy ta có:
$x^{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\geq 4\sqrt[4]{\frac{x^2}{8}}=\frac{4}{\sqrt[4]{8}}\sqrt{x}<=>x^2+\frac{3}{2}\geq \frac{4}{\sqrt[4]{8}}\sqrt{x}$
$y^2+\frac{3}{2}\geq \frac{4}{\sqrt[4]{8}}\sqrt{y}$
$<=> x^2+y^2+3\geq \frac{4}{\sqrt[4]{8}}A<=>4\geq\frac{4}{\sqrt[4]{8}}A<=>\sqrt[4]{8}\geq A<=>maxA=\sqrt[4]{8}$
Dấu "=" xảy ra $<=>x=y=\frac{1}{\sqrt{2}}$
Mặt khác: Vì $x^2+y^2=1<=>0<x,y<1<=> \sqrt{x} \geq x^2, \sqrt{y} \ geq y^2$
nên $A \ geq x^2+y^2=1<=> minA =1$
Dấu "=' xảy ra $<=> x=1, y=0$ hoặc $x=0, y=1$