I am not quite sure what the question is asking.

The absolute value of $\operatorname{cov}(X,Y)$, the covariance of $X$ and $Y$ is no larger than $\sigma_X\sigma_Y$ which is the geometric mean of the variances of $X$ and $Y$. Since $$\min\{\sigma^2_X,\, \sigma^2_Y\} \leq \sigma_X\sigma_Y \leq \max\{\sigma^2_X,\, \sigma^2_Y\},$$ it is certainly possible for the covariance to exceed $\min\{\sigma^2_X,\, \sigma^2_Y\}$. In other words, the question

Is it not true in general that $\big|\operatorname{cov}(X, Y)\big| \leq \min\{\sigma^2_X,\, \sigma^2_Y\}$?

has the answer that the desired relationship is not always feasible. Consider, for example, $X$ and $Y$ having variances $6^2$ and $8^2$ respectively and correlation coefficient $5/6$. Then, $$\operatorname{cov}(X, Y) = \rho\sigma_X\sigma_Y = \frac 56 \times 6 \times 8 = 40 > \min \{\sigma^2_X,\, \sigma^2_Y\} = 36.$$

Also $\displaystyle \frac{\sigma_X}{\sigma_Y} = \frac{6}{8} < \rho = \frac 56.$

With a smaller correlation coefficient $\frac 34$, we have that the covariance is $36$, same as $\sigma_X^2$, and of course $\displaystyle \frac{\sigma_X}{\sigma_Y} = \frac{6}{8} = \rho$.

I am not quite sure what the question is asking.

The absolute value of $\operatorname{cov}(X,Y)$, the covariance of $X$ and $Y$ is no larger than $\sigma_X\sigma_Y$ which is the geometric mean of the variances of $X$ and $Y$. Since $$\min\{\sigma^2_X,\, \sigma^2_Y\} \leq \sigma_X\sigma_Y \leq \max\{\sigma^2_X,\, \sigma^2_Y\},$$ it is certainly possible for the covariance to exceed $\min\{\sigma^2_X,\, \sigma^2_Y\}$. In other words, the question

Is it not true in general that $\big|\operatorname{cov}(X, Y)\big| \leq \min\{\sigma^2_X,\, \sigma^2_Y\}$?

has the answer that the desired relationship is not always feasible.

Consider, for example, $X$ and $Y$ having variances $6^2$ and $8^2$ respectively.

• Suppose that the correlation coefficient $\rho$ is $\displaystyle \frac 56$.
  Then, $\displaystyle\operatorname{cov}(X, Y) = \rho\sigma_X\sigma_Y = \frac 56 \times 6 \times 8 = 40 > \min \{\sigma^2_X,\, \sigma^2_Y\} = 36,$ and $\displaystyle \frac{\sigma_X}{\sigma_Y} = \frac{6}{8} < \rho = \frac 56.$

• With a smaller correlation coefficient $\displaystyle \frac 34$, we have that the covariance is $36$, same as $\sigma_X^2$, and of course $\displaystyle \frac{\sigma_X}{\sigma_Y} = \frac{6}{8} = \rho$.

• If $\rho$ were even smaller, say $\displaystyle\rho = \frac 12$, the covariance is $24$ which is smaller than $\sigma_X^2$.

So it would appear that the OP is asking whether most real-life data sets that people have encountered (with $\sigma_X < \sigma_Y$) happen to have correlation coefficients that do not exceed $\displaystyle\frac{\sigma_X}{\sigma_Y}$ in magnitude.