回復 6# Christina 的帖子置
第三題：
\(\displaystyle f\left(x\right)=\left|4x-3a\right|+\left|5x-4a\right|=4\left|x-\frac{3a}{4}\right|+5\left|x-\frac{4a}{5}\right|\)
\(\displaystyle =4\left(\left|x-\frac{3a}{4}\right|+\left|x-\frac{4a}{5}\right|\right)+\left|x-\frac{4a}{5}\right|\)
\(\displaystyle \geq 4\left|\left(x-\frac{3a}{4}\right)-\left(x-\frac{4a}{5}\right)\right|+\left|x-\frac{4a}{5}\right|\)
\(\displaystyle =\frac{\left|a\right|}{5}+\left|x-\frac{4a}{5}\right|\)
可知，當 \(\displaystyle x=\frac{4a}{5}\) 時， \(f\left(x\right)\) 有最小值為 \(\displaystyle \frac{\left|a\right|}{5}\)
\(\displaystyle\frac{\left|a\right|}{5}\geq a^2\Rightarrow \frac{\left|a\right|}{5} \geq \left|a\right|^2\Rightarrow \frac{-1}{5}\leq a\leq \frac{1}{5}\)
故，當 \(\displaystyle\frac{-1}{5}\leq a\leq \frac{1}{5}\) 時，\(f\left(x\right)\geq a^2\) 恆成立。
