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kuing
Posted 2021-3-11 15:47
由条件得 `\abs{\bm d}=2` 且 `\abs{2\bm a+\bm c}=\abs{\bm a+2\bm c}`,所以
\begin{align*}
\text{原式}&=2\abs{1-\lambda}+\abs{\bm a+2\bm c}+\abs{2\bm a-2\bm c-2\lambda\bm d}\\
&\geqslant2\abs{1-\lambda}+\abs{3\bm a-2\lambda\bm d}\\
&=2\abs{1-\lambda}+\abs{(3-2\lambda)\bm a+2\sqrt3\lambda\bm b}\\
&=2\abs{1-\lambda}+\sqrt{16\lambda^2-12\lambda+9}\\
&=2\abs{1-\lambda}+\sqrt{\left( \frac14+\frac34 \right)\left( \left( 4\lambda-\frac32 \right)^2+\frac{27}4 \right)}\\
&\geqslant2(1-\lambda)+2\lambda-\frac34+\frac94\\
&=\frac72,
\end{align*}取等懒得算了,略…… |
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