$\mathbf{E}\left[\frac{(U_1+c)^2}{\max((U_1+c)^2, U_2^2)} \right] \ge \mathbf{E}\left[\frac{U_2^2}{\max((U_1+c)^2, U_2^2)} \right]$

Question

We consider two i.i.d. random variables $U_1$ and $U_2$ such that $\mathbf{E}[U_1] = \mathbf{E}[U_2] = 0$ and $\textrm{Var}[U_1] = \textrm{Var}[U_2] < \infty$.

Prove that for any $c > 0$ the following inequality holds: $\mathbf{E}\left[\frac{(U_1+c)^2}{\max((U_1+c)^2, U_2^2)} \right] \ge \mathbf{E}\left[\frac{U_2^2}{\max((U_1+c)^2, U_2^2)} \right]$.

Additional constraints:

1. The probability distribution of $U_1$ and $U_2$ is symmetric around zero.
2. The probability distribution of $U_1$ and $U_2$ has its mode at zero?

---

My attempt

The original problem might be ill-posed in the sense that it only works for a restricted class of probability distributions, but I cannot seem to be able to prove it even when $U_1$ and $U_2$ follow the standard normal distribution, although the experimental results indicate that the inequality might hold for this particular case.

I have tried splitting the integral from the expected value to get rid of the 'max' term:

$ \int_{(U_1+c)^2 > U_2^2} p(u_1) p(u_2) du_1 du_2 + \int_{(U_1+c)^2 \le U_2^2} \frac{(u_1+c)^2}{u_2^2} p(u_1) p(u_2) du_1 du_2 \ge \int_{(U_1+c)^2 > U_2^2} \frac{u_2^2}{(u_1+c)^2} p(u_1) p(u_2) du_1 du_2 + \int_{(U_1+c)^2 \le U_2^2} p(u_1) p(u_2) du_1 du_2 $

Unfortunately, this does not make the problem easier for me, even in the standard normal case.

I can only show the following more general result:

$\left| \mathbf{E}\left[\frac{(U_1+c)^2}{\max((U_1+c)^2, U_2^2)} \right] - \mathbf{E}\left[\frac{U_2^2}{\max((U_1+c)^2, U_2^2)} \right] \right| \le 1$.

Proof: $ \; \max((U_1+c)^2, U_2^2) = \max(|U_1+c|, |U_2|)^2 = \left( \frac{|U_1+c| + |U_2| + ||U_1+c| - |U_2||}{2} \right)^2 \overset{AM-GM}{\ge} (\sqrt{|(U_1+c)^2 - (U_2)^2|})^2 = |(U_1+c)^2 - (U_2)^2| $

$\left| \mathbf{E}\left[\frac{(U_1+c)^2}{\max((U_1+c)^2, U_2^2)} \right] - \mathbf{E}\left[\frac{U_2^2}{\max((U_1+c)^2, U_2^2)} \right] \right| = \left| \mathbf{E} \left[\frac{(U_1+c)^2 - U_2^2}{\max((U_1+c)^2, U_2^2)} \right] \right| \le \mathbf{E} \left[ \frac{|(U_1+c)^2 - U_2^2|}{\max((U_1+c)^2, U_2^2)} \right] \le \mathbf{E} \left[ \frac{|(U_1+c)^2 - U_2^2|}{|(U_1+c)^2 - U_2^2|} \right] = \mathbf{E} [1] = 1 \qquad \textrm{q.e.d.} $

---

I have run some simple Monte Carlo experiments in R and this results seems to hold at least in the case of normally or uniformly distributed variables. Here is some sample code:

N <- 1e6
c <- 1 

u1 <- rnorm(N) + c # runif(N) + c
u2 <- rnorm(N)     # runif(N)

yy <- pmax(u1 * u1, u2 * u2)
mean((u1 * u1) / yy) - mean((u2 * u2) / yy) # >= 0

After running more experiments and giving some more thought to the problem, I have come to the conclusion that additional constraints on the probability distribution of the random variables might be necessary for the above inequality to hold.

I have tried running the above simulation for a positively skewed normal distribution via the rsnorm function in the fGarch R package. The inequality does not seem to hold for positively skewed distributions with expected value zero, but it does on the other hand hold for symmetric and negatively skewed distributions with expected value zero.

It might also be the case that the probability distribution of $U_1$ and $U_2$ needs to have its / a mode at zero. I have tried the experiment for a simple mixture of two normal distributions, specifically $\frac{1}{2} (\mathcal{N}(-5, 1) + \mathcal{N}(5, 1))$, with $c = \frac{1}{10}$, and it seems that the inequality does not hold, although the difference is small enough that it might be due to numerical errors.