Prove radiance contributions of deeper paths fade away
If you are asking whether:

$$
\int fr(...) cos\theta_1Le_1 &gt; \int fr(...)cos\theta_1 fr(...)cos\theta_2 Le_2
$$

(simplified terms from the render equation you posted above), I don't think you could prove this. Le_2 could just be much greater than Le_1. 

For example you could have a set of mirrors which only hit a light source after two bounces. Then all the light would come from the second integral.



EDIT: Updating answer after your response.

If you are trying to prove that, ignoring the layout and lights of the scene, deeper paths contribute less to the overall equation, then you could try ignoring Le and assuming it is always 1. You are then trying to prove:

$$
\int_\Omega fr(...)cos\theta_1  &gt; \int_{\Omega^2}fr(...)cos\theta_1fr(...)cos\theta_2
$$
or
$$
\int_{\Omega_1} fr(...)cos\theta_1 dw_1 &gt; \int_{\Omega_1} fr(...)cos\theta_1\int_{\Omega_2} fr(...)cos\theta_2 dw_2 dw_1
$$

We also know that if the fr is energy conserving:

$$
0 \leq \int fr(...)  \leq 1
$$

And if 0 &lt; cos &lt; 1 we can say that:

$$
0 \leq \int fr(...)cos\theta \leq 1
$$

Substituting this knowlege into the equation into the inner integration of the right hand side doesnt give us much, as it could always be equal to 1 which would give us: 

$$
\int_{\Omega_2} fr(...)cos\theta_2 dw_2  = 1
$$
$$
\int_{\Omega_1} fr(...)cos\theta_1 dw_1 &gt; \int_{\Omega_1} fr(...)cos\theta_1 dw_1
$$

which disproves your idea. However if your bxdf is never perfectly energy conserving, eg:

$$
0 \leq \int fr(...)  &lt; 1
$$

I think subsitituting this in does prove the above as we get something like:

$$
0 \leq n(w) &lt; 1
$$
$$
\int_{\Omega_1} fr(...)cos\theta_1 dw_1 &gt; \int_{\Omega_1} fr(...)cos\theta_1 n(w)dw_1
$$

As each term can only be less than the integral on the left.

Hopefully my math is okay here I might have made a mistake. You may need to clean it up and do some more formal proofs. I am sorry if this still doesnt help answer your question.