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Let $ X $ and $ Y $ be two random variables that satisfy $ E[X^2] + E[Y^2] < \infty $. Prove that $$ E[\sqrt{X^2 + Y^2}] \geq \sqrt{(E[X])^2 + (E[Y])^2}. $$ Based on the inequality above, construct the corresponding elementary inequality for a sequence of real numbers.

I can do the second part: The inequality $ E[\sqrt{X^2 + Y^2}] \geq \sqrt{(E[X])^2 + (E[Y])^2} $ suggests a corresponding inequality in the context of real numbers. For real numbers $ x_1, x_2, \dots, x_n $ and $ y_1, y_2, \dots, y_n $ , we can derive the following elementary inequality:

$$ \frac{1}{n} \sum_{i=1}^{n} \sqrt{x_i^2 + y_i^2} \geq \sqrt{\left( \frac{1}{n} \sum_{i=1}^{n} x_i \right)^2 + \left( \frac{1}{n} \sum_{i=1}^{n} y_i \right)^2 }. $$

This is the corresponding elementary inequality for a sequence of real numbers.

I'm stuck at the first part of the problem. I have tried to apply Jensen's inequality which states that that for a convex function $ f $ and a random variable $ Z $, $$ E[f(Z)] \geq f(E[Z]). $$ However,in this case, the function $ f(z) = \sqrt{z} $ is concave.
Also, I have proved that $f(x,y)=\sqrt{x^2+y^2}$ is convex in $\mathbb{R}^2.$
Any help for the proof of the first part's inequality I would be very grateful!

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    $\begingroup$ Since you proved that $f(x,y)=\sqrt{x^2+y^2}$ is convex, what you want to prove is a direct consequence of the multivariate version of Jensen’s inequality. $\endgroup$ Commented Sep 21, 2024 at 15:48

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$$ \begin{align} & \;\;\;\;\,(EX)^2+(EY)^2 \\&= E[X\cdot EX+Y\cdot EY] \\&\le E\left[\sqrt{X^2+Y^2}\cdot \sqrt{(EX)^2+(EY)^2}\right]\tag1 \\&=E[\sqrt{X^2+Y^2}]\cdot \sqrt{(EX)^2+(EY)^2} \end{align} $$ The inequality in $(1)$ is the result of applying Cauchy-Schwarz to the inner product of $(X,Y)$ with $(EX,EY)$.

Letting $r=\sqrt{(EX)^2+(EY)^2}$, the above proves that $r^2\le E[\sqrt{X^2+Y^2}]\cdot r$. Conclude by dividing both sides by $r$.

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$\sqrt {a^{2}+b^{2}}=\sup \{ac+bd: c^{2}+d^{2}=1\}$. So, RHS is $\sup \{cEX+dEY: c^{2}+d^{2}=1\}$. So RHS $=\sup \{E(cX+dY): c^{2}+d^{2}=1\}$. For any $c,d $ with $c^{2}+d^{2}=1$ we have $E(cX+dY) \le E\sqrt {X^{2}+Y^{2}}$ from which your inequality follows.

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