Hong Kong Preliminary Selection Contest

It suffices to minimise $2|x+1|+4|x-5|+4\left|x-\frac{7}{2}\right|+|x-11|$ (which is 2 times the given expression), i.e. to find the minimum total distance from $x$ to the 11 numbers: $$-1,-1,\frac{7}{2},\frac{7}{2},\frac{7}{2},\frac{7}{2},5,5,5,5,11.$$ By the triangle inequality, we have $|x-a|+|x-b|\ge |(x-a)+(b-x)|=b-a$ for any $a\le b$, and the equality holds whenever $x$ lies between $a$ and $b$. From this, we try to pair up the numbers as follows: $(-1,11)$, $(-1,5)$, $(\frac{7}{2},5)$, $(\frac{7}{2},5)$, $(\frac{7}{2},5)$, leaving a single number $\frac{7}{2}$. From the above discussion, the total distance from $x$ to each pair is minimised when $x$ lies between them. Hence we can achieve the minimum by minimising the distance from $x$ to $\frac{7}{2}$ while making sure that $x$ lies between each pair of numbers. Clearly, this can be achieved by taking $x=\frac{7}{2}$. It follows that the answer is $\left|\frac{7}{2}+1\right|+2\left|\frac{7}{2}-5\right|+2\times \frac{7}{2}-7+\left|\frac{7}{2}-11\right|=\frac{45}{4}$.