7. If real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=3, x+2 y-2 z=4$, then $z_{\text {max }}+z_{\text {min }}=$
Answer $-\frac{16}{9}$.
Analysis By the Cauchy-Schwarz inequality, we have $\left(x^{2}+y^{2}\right)\left(1+2^{2}\right) \geq(x+2 y)^{2}$. From the given, we have $x^{2}+y^{2}=3-z^{2},(x+2 y)^{2}=(4+2 z)^{2}$, so we get $5\left(3-z^{2}\right) \geq(4+2 z)^{2}$. Simplifying, we get $9 z^{2}+16 z+1 \leq 0$, which means $z_{\max }, z_{\min }$ are the roots of the equation $9 z^{2}+16 z+1=0$. By Vieta's formulas, we get $z_{\max }+z_{\min }=-\frac{16}{9}$.