An upper bound for the ratio of one random variable to the square root of the sum of squares of random variables Let $X_1, X_2, X_3$ be positive, iid, non-degenerate random variables with finite variances.
I wish to show that
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] < \dfrac{2}{\sqrt{3}}\text{.}$$
My efforts. We can start off by observing that
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] = 2 \cdot \mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right]$$
so it is sufficient to prove that
$$\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] < \dfrac{1}{\sqrt{3}}\text{.}$$
I am not sure how to go about this step.
One could try to handwave replacing $X_2$ and $X_3$ with $X_1$, but I can't think of a way of theoretically justifying that (it's probably a bad idea).
 A: Consider the distribution of $(X_1, X_2, X_3)$. Because $X_1, X_2, X_3$ are iid, it follows that they are exchangeable, hence the distribution of $(X_1, X_2, X_3)$ is the same as $(X_2, X_1, X_3)$.
By linearity of expectation, we have that
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] = \mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] + \mathbb{E}\left[\dfrac{X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right]\text{.}$$
The two terms of the right-hand side of the above equation are equal by exchangeability, hence we know that
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] = 2 \cdot \mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right]$$
so it suffices to prove that
$$\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] < \dfrac{1}{\sqrt{3}}\text{.}$$
By Cauchy-Schwarz, observing that $\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} > 0$ with probability $1$, and therefore $\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] > 0$, we have that
$$\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}}\right] = \mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \cdot 1 \right] \leq \sqrt{\mathbb{E}\left[\dfrac{X_1^2}{X_1^2 + X_2^2 + X_3^2} \right]}\text{.}$$
Observe that
$$\mathbb{E}\left[\dfrac{X_1^2 + X_2^2 + X_3^2}{X_1^2 + X_2^2 + X_3^2} \right] = 1 = \sum_{i=1}^{3}\mathbb{E}\left[\dfrac{X_i^2}{X_1^2 + X_2^2 + X_3^2} \right]\text{.}$$
Exploiting exchangeability again, we have that
$$1 = \sum_{i=1}^{3}\mathbb{E}\left[\dfrac{X_i^2}{X_1^2 + X_2^2 + X_3^2} \right] = 3 \cdot \mathbb{E}\left[\dfrac{X_1^2}{X_1^2 + X_2^2 + X_3^2} \right]$$
hence
$$\mathbb{E}\left[\dfrac{X_1^2}{X_1^2 + X_2^2 + X_3^2} \right] = \dfrac{1}{3}$$
hence
$$\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] \leq \dfrac{1}{\sqrt{3}}\text{.}$$
However, equality holds if and only if $\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}}$ is constant (i.e., proportional to $1$) almost surely, so we can drop the equality.
