10. Real numbers $x, y$ satisfy $x^{2}+y^{2}=20$, then the maximum value of $x y+8 x+y$ is

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Rong Ge 42.
Analysis Note $x y \leq \frac{1}{4} x^{2}+y^{2}, 8 x \leq x^{2}+16, y \leq \frac{1}{4} y^{2}+1$, adding these three inequalities yields
$$
x y+8 x+y \leq \frac{5}{4}\left(x^{2}+y^{2}\right)+17=42
$$

and when $x=4, y=2$, the equality can be achieved, so the maximum value of $x y+8 x+y$ is 42.
Alternatively, the Cauchy inequality can be used directly
$$
(x y+8 x+y)^{2} \leq\left(x^{2}+8^{2}+y^{2}\right)\left(y^{2}+x^{2}+1^{2}\right)=84 \cdot 21=42^{2}
$$

yielding the maximum value of 42.