Prove $\log_{\frac{1}{4}} \frac{8}{7}> \log_{\frac{1}{5}} \frac{5}{4}$ 
Prove $$\log_{\frac{1}{4}} \frac{8}{7}> \log_{\frac{1}{5}} \frac{5}{4}$$

How to prove without a computer?

My second attempt, hopefully not flawed this time. (Although I think that there must be a simpler way to do this.)
I will use $\ln$ to denote natural logarithm. (Some people use $\log$)
The inequality
$$\log_{\frac{1}{4}} \frac{8}{7}> \log_{\frac{1}{5}} \frac{5}{4}$$
can be rewritten as
$$\frac{\ln\frac87}{\ln\frac14} > \frac{\ln\frac54}{\ln\frac15}.$$
This is equivalent to the inequality
$$\ln\frac87 \ln \frac15 > \ln \frac54 \ln\frac14$$
(we have multiplied both sides of the inequality by $\ln\frac14$ and $\ln\frac15$. So we have multiplied the inequality by a negative number twice - as a final result the inequality signs remains unchanged.)
This leads to equivalent inequalities
$$-\ln 5 \ln\frac87 > -\ln 4 \ln \frac54\\
\ln 4 \ln \frac54 > \ln 5 \ln\frac87. \tag{*}$$
So the above inequality $(*)$ is what we want show.
We know that $\ln(1+x)<x$ and $\ln(1+x)>\frac{x}{1+x}$ for $x>0$.
(Both these inequalities can be verified simply by taking derivative of functions $\ln(1+x)-x$ and $\ln(1+x)-\frac{x}{1+x}$ and showing that the derivative is negative/positive for $x>0$.
See WolframAlpha: 1, 2. Of course, there can be other ways how to prove this.
Both these properties are mentioned at Wikipedia.)
Now if we use one of these inequalities for $x=\frac17$, we get $\ln\frac87 < \frac17$, which gives
$$\ln\frac87\ln 5 < \frac17\ln5.$$
For $x=\frac14$ we have $\ln\frac54 > \frac{1/4}{1+1/4} = \frac{1/4}{5/4} = \frac15$ and thus
$$\ln \frac54 \ln 4 > \frac15 \ln 4.$$
Now we may notice that the following inequalities are equivalent to each other:
$$\frac15 \ln 4 > \frac17\ln5 \tag{**}\\
7\ln 4 > 5\ln 5\\
\ln(4^7) > \ln{5^5}\\
4^7 > 5^5\\
$$
The last inequality is true, which can be seen by evaluating $4^7=16384$ and $5^5=3125$. (And this can be shown by hand, too, since
$4^7=2^{14}=16\cdot2^{10}=16\cdot1024 > 16000$ and $5^5=125\cdot 25 < 128 \cdot 25 =  32 \cdot (4\cdot 25) = 3200$.) 
This proves $(**)$.
So we have shown the following chain of inequalities
$$\ln \frac54 \ln 4> \frac15 \ln 4 > \frac17\ln5 > \ln\frac87\ln 5.$$
If we compare the leftmost and the rightmost expression, this is precisely $(*)$.
