856

已知等比数列 ${\displaystyle \{a_n\}}$ 满足 ${\displaystyle a_n>0}$ 且 ${\displaystyle a_1{a}_2{a}_3+2a_2^2+a_2+a_3-a_4=1}$, 求 ${\displaystyle a_1}$ 的取值范围.

记 ${\displaystyle \{a_n\}}$ 公比为 ${\displaystyle q>0}$.则条件式等价于$$\begin{array}{r}f(q)=(a_1^3-a_1)q^3+a_1(2a_1+1)q^2+a_{1}q-1=0,\end{array}$$
其中 ${\displaystyle f}$ 的定义域为 ${\displaystyle (0,+\mathrm{\infty })}$.而 ${\displaystyle f^{\prime }(q)=[3(a_1+1)q+1][(a_1-1)q+1]}$.

• 若 ${\displaystyle a_1\ge 1}$，则 ${\displaystyle f^{\prime }(q)>0}$.又 ${\displaystyle f(0)=-1<0,{\displaystyle \underset{q\to +\mathrm{\infty }}{lim}f(q)=+\mathrm{\infty }}}$,故 ${\displaystyle f(q)}$ 存在唯一零点, 符合题意.
• 若 ${\displaystyle 0<a_1<1}$，则 ${\displaystyle f(q)}$ 有唯一极大值点 ${\displaystyle q_0={\displaystyle \frac{1}{1-a_1}>0}}$，因此要求$$\begin{array}{r}f\left(\frac{1}{1-a_1}\right)=-\frac{a_1^2-3a_1+1}{(a_1-1{)}^2}\ge 0\Rightarrow \frac{3-\sqrt{5}}{2}\le a_1<1.\end{array}$$

综上，${\displaystyle a_1\ge {\displaystyle \frac{3-\sqrt{5}}{2}}}$.