Show df $> m-1$ for Welch two sample t-test

Question

I am trying to verify that the degrees of freedom $k$ for a Welch two sample $t$-test with unequal variances has $k > m-1$. It is quite clear for the case when they have equal variances.

The formula for df is $k =\dfrac{\left(\frac{S_1^2}{m} + \frac{S_2^2}{n}\right)^2}{ \frac{(S_1^2/m)^2}{m-1}+\frac{(S_2^2/n)^2}{n-1}}$.

I have tried a few difference approaches to manipulate this but I haven't had much success. Any help would be appreciated. Thanks.

Answer 1

Comment continued: Special case where $n = 10, m = 20,$ and (depending on ratio of sample variances) $0 < c < 1.$ Plot in R statistical software.

c = seq(.001, .999, by=.001);  n = 10;  m=20
denom=(1-c)^2*(n-1)+c^2*(m-1)
k = (n-1)*(m-1)/denom
plot(c, k, type="l", lwd=2)
  abline(v=0:1, col="green2");  abline(h=c(9, 19, 28), col="red")