GTNN
\(x^2+y^2=1=\left(x+y\right)^2-2xy\Rightarrow2xy=\left(x+y\right)^2-1\)
\(x;\text{ }y\ge0\Rightarrow x+y=\sqrt{x^2+y^2+2xy}\ge\sqrt{1+2xy}\ge1\)
\(A^2=2+2\left(x+y\right)+2\sqrt{\left(1+2x\right)\left(1+2y\right)}\)
\(=2+2\left(x+y\right)+2\sqrt{1+2\left(x+y\right)+4xy}\)
\(=2+2\left(x+y\right)+2\sqrt{1+2\left(x+y\right)+2\left(x+y\right)^2-2}\)
\(=2+2t+2\sqrt{2t^2+2t-1}\text{ }\left(t=x+y\ge1\right)\)
\(\ge2+2+2\sqrt{2.1^2+2.1-1}\)
\(=4+2\sqrt{3}\)
\(\Rightarrow A\ge\sqrt{4+2\sqrt{3}}=1+\sqrt{3}\)
Dấu bằng xảy ra khi \(x+y=1\Leftrightarrow xy=0\Leftrightarrow\left(x;y\right)=\left(1;0\right);\left(0;1\right)\)
GTLN
Với 2 số thực bất kì, ta luôn có: \(\left(a+b\right)^2=2\left(a^2+b^2\right)-\left(a-b\right)^2\le2\left(a^2+b^2\right)\)
\(A^2\le2\left(1+2x+1+2y\right)=4+4\left(x+y\right)\le4+4\sqrt{2\left(x^2+y^2\right)}=4+4\sqrt{2}\)
\(\Rightarrow A\le\sqrt{4+4\sqrt{2}}\)
Dấu bằng xảy ra khi 2 biến bằng nhau.
