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?
 A: Here is another argument using power-series-based computations of $\log$, but slightly different in its details to Ivan's.
$$\log_{1/4} 8/7 = -\log_4 8/7 = \log_4 7/8 =  \log_4 (1 - 1/8),$$
while 
$$\log_{1/5} 5/4 = -\log_5 5/4 = \log_5 4/5 = \log_5 (1 - 1/5).$$
So the problem reduces to showing that
$$\log_4(1-1/8) >\log_5 (1- 1/5).$$
This is slightly delicate, because both numbers of which we are taking the $\log$ are close to $1$, so both sides are close to $0$, and we have to estimate how close.  
Recall that $$\ln (1- x) = - x - x^2/2 - x^3/3 - \cdots$$
when $|x| < 1$,
and that $\log_a (1-x) = (1/\ln a)\ln (1-x)$.
So we have to prove
$$\ln 4 \, (1/5 + 1/50 + \cdots ) > \ln 5 \,(1/8 + 1/128 + \cdots).$$
Now clearly
$$(1/5 + 1/50 + \cdots ) > 8/5 (1/8 + 1/128 + \cdots),$$
so it suffices to show that 
$$(8/5)\ln 4 > \ln 5,$$
i.e. that $$\ln 4^8 > \ln 5^5,$$
i.e. that $$65536 > 3125,$$
which is true.
A: 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 $(*)$.
A: $\newcommand{\Log}{\operatorname{Log}}$I will use approximations by power series. It is easy to show that the inequality is equivalent to:$\Log(3.5) \Log(5) > \Log(4)^2$. On the other hand consider the power series expansion for $\Log(x)$ around $3.5$.It is: $$\Log(3.5)+\frac{2}{7}(x-3.5)-\frac{2}{49}(x-3.5)^2+O(x-3.5)^3$$ For $x=5$ we get: $\Log(5)>33/98+\Log[3.5]$
On the other hand for $x=4$ the linear term gives: $\Log(4)<1/7+\Log(3.5)$. Now we aim to prove the following:$$
\Log(3.5) \Log(5) > \Log(3.5)*(33/98 + \Log(3.5)) > (1/7 + \Log(3.5))^2 >
 \Log(4)^2$$
The first and last are proved already by the approximations. We need the middle. It is equivalent to the positivity of:$$
\Log(3.5)*(33/98 + \Log(3.5)) - (1/7 + \Log(3.5))^2
=-(1/49) + 5/98 \Log(3.5)
=(1/49)(5/2\Log(3.5)-1)
$$
The last is positive because $\Log(3.5)>1>2/5$.
