solved Why geometric mean get a same relative answer rather than arithmetic mean? I encountered this problem while studying SPEC(System Performance Evaluation Cooperative) test for CPU measurment. But I think this is a mathematical problem, not a computer science.
SPEC uses a set of programs to test the CPU, and each program returns an execution time. The SPECratio for each program is the inverse of execution time(CPU) / execution time(running in reference computer, such as a standard machine). The final mark SPEC given by SPEC is the geometric mean of all SPECratio.
They give an explantation why geometry instead of arithmetic should be used.

When comparing two computers using SPECratios, use the geometric
mean so that it gives the same relative answer no matter what computer is used to
normalize the results. If we averaged the normalized execution time values with an
arithmetic mean, the results would vary depending on the computer we choose as the
reference.

I don’t know why they say that. Is it related to the geometric mean feature? Can someone help? ：）
Below is the mark of Intel Core i7 920 (picture from Computer Organization and Design the Hardware/Software Interface)

note: there'are not any functional dependence, just add or product, then divide
when comparing 2 marks, the form using geometric mean can convert to form
to $\frac{t1*t2*...*tn}{t01*t02*...*t0n}$ vs $\frac{t1'*t2'*...*tn'}{t01*t02*...*t0n}$
The factor $t01*t02*...*t0n$ will cancel out, so we will get the same relative answer
When using arimetric mean, it becomes
$\frac{t1}{t01} + \frac{t2}{t02} + ... + \frac{tn}{t0n}$  vs $\frac{t1'}{t01} + \frac{t2'}{t02} + ... + \frac{tn'}{t0n}$
And the effect of reference computer can not be ignored.
thanks help for @lenobloy
 A: We have a reference computer ($0$) and several other computers ($1,2\cdots$). Each one executes several tests $j=1,2 \cdots J$, let $t_i^{(j)}$ be the time that it takes the computer $i$ to accomplish task $j$
To compare computer $1$ against $2$, one could compare the arithmetic mean or the geometric mean, i.e
$$ \bar t_i = \frac{1}{J} \sum_{j=1}^J t_i^{(j)}$$
or
$$ {\tilde t_i} = \sqrt[J]{ \prod_{j=1}^J t_i^{(j)}}$$
It's not clear which is to be preferred. However, suppose that instead of being given the "absolute" times $t_i^{(j)}$ one is given the "relative" times, i.e., the ratio against the times from some reference computer:
$$r_i^{(j)} = \frac{t_i^{(j)}}{t_0^{(j)}}$$
In this case, if one uses the geometric mean for $r_i^{(j)} $ the computer $i$ will be judged better than $i$ if and only if its (geometric mean) absolute time is also better.
Because the reference times ${t_0^{(j)}}$ will cancel out in the comparison.
This will not be the case if the arithmetic mean is used. In this case the comparison will depend on the reference times ${t_0^{(j)}}$, hence it could happen that computer $1$ performs better than computer $2$ if using the ratio against some computer $0$, but whn using another reference computer the result is the opposite.
This is clearly undesirable. Hence, the geometric mean is to be preferred.
