Let $a$ and $b$ be real numbers.  Then $(a - b)^2 \ge 0,$ which is equivalent to
\[ab \le \frac{a^2 + b^2}{2}.\](This looks like AM-GM, but here, we want to show that it holds for all real numbers, not just nonnegative real numbers.)

Setting $a = x$ and $b = \sqrt{1 - y^2},$ we get
\[x \sqrt{1 - y^2} \le \frac{x^2 + 1 - y^2}{2}.\]Setting $a = y$ and $b = \sqrt{1 - x^2},$ we get
\[y \sqrt{1 - x^2} \le \frac{y^2 + 1 - x^2}{2}.\]Therefore,
\[x \sqrt{1 - y^2} +y \sqrt{1 - x^2} \le \frac{x^2 + 1 - y^2}{2} + \frac{y^2 + 1 - x^2}{2} = 1.\]Since $f(1,0) = 1,$ the maximum value is $\boxed{1}.$
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Find the maximum value of
\[f(x,y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2},\]where $-1 \le x,$ $y \le 1.$
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
1
Find the maximum value of
\[f(x,y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2},\]where $-1 \le x,$ $y \le 1.$
test