Show/Prove that $F_{\alpha,n,m} =1/F_{1-\alpha,m,n}$ Show/Prove that $F_{\alpha,n,m} = \frac{1}{F_{1-\alpha,m,n}}$
The distributions I'm working with are the Fisher distribution and the Chi-square distribution.
I can prove that n and m switch for the F distribution as follows:
$$X = \frac{ \frac{A}{n} } { \frac{B}{m} }$$
where A and B follow $\chi^2$ distributions, are independent, and have n, m degrees of freedom respectively.  Then the following holds:
$$\frac{1}{X} = \frac{ \frac{B}{m} } { \frac{A}{n} }$$
My question is how do I show that the $F_{\alpha}$ part goes to $F_{1-\alpha}$
Is there a way to do this all at once or do I need to show the degrees of freedom switch separately from the $1-\alpha$ part of the proof?
 A: If you already established that
$$
1/F_{(m,n)} = F_{(n,m)}, 
$$
by the reciprocal of the chi square distributions,
then let $X\sim F_{(m,n)}$ hence $1/X\sim F_{(n,m)}$, so
\begin{align}
\alpha =& P(X\le F_{\alpha, (m,n)})\\
        =& P(1/X \ge 1/F_{a, (m,n)} )\\
        =& 1 - P(1/X \le 1/F_{a, (m,n)} )\\
         =&1-P( F_{(n,m)} \le 1/F_{a, (m,n)})\\
         =&1-F_{(n,m)}(1/F_{a, (m,n)})    
\end{align}
thus,
\begin{align}
1-\alpha =& F_{(n,m)}(1/F_{a, (m,n)})\\
F_{1-\alpha, (n,m)}         =& 1/F_{a, (m,n)}. 
\end{align}
A: This relationship is used to get lower tail probabilties of F-distributions.
Most tables of F-distributions give only upper tail probabilities.

The succinct answer to your main question is as follows: When you take reciprocals, the sense of the inequality is changed:
  $P(X \le c) = P(1/X \ge 1/c).$

Statistical software usually gives the complete PDF of a distribution.
Suppose $X \sim \mathsf{F}(5, 10)$ and you want to cut off the lower 5%
of the probability in this distribution. That is, you seek $c$ such that $P(X \le c) = 0.05.$
In R statistical software quantile 0.05 of $X$ is $c = 0.2112$ to four places.
qf(.05, 5, 10)       # in R 'qf' is the inverse CDF or quantile function
## 0.2111904
pf(.2111904, 5,10)   # in R 'pf' is the CDF
## 0.04999999

Your result shows that this is the same as cutting 5% of the probability
of the probability in the distribution $\mathsf{F}(10, 5)$ in the upper tail
and taking the reciprocal. 
That is, $Y = 1/X \sim \mathsf{F}(10, 5),$ so that $P(Y \ge d) = .95$ and using
$c=1/d$ as the lower 5% cutoff of $X.$ Here $d = 4.735,$ which you should verify
from the F-table in your textbook. Also, $c = 1/d = 0.2112$ cuts 5% from
the lower tail of $\mathsf{F}(5, 10),$ as previously found directly from R software.
 d = qf(.95, 10, 5);  d;  1/d
 ## 4.735063   # d = 4.735 is the 95th quantile of F(10, 5)
 ## 0.2111904  # c = 1/d = 0.2112 is the 5th quantile of F(5, 10)

Note: In the notation $F_{\alpha, n, m},$ the probability $\alpha$ is
not a quantile; it is an upper-tail probability.

The figure below shows the density function of $\mathsf{F}(5,10).$
The areas beneath the density to the left of 0.2112 and to the right of
3.326 are both 5%. You can find the right-hand cutoff from printed
tables of the F-distribution.

