Proving $E[\max(X^2,Y^2)]\le 1+\sqrt{1-\rho^2}$

Question

The question is

Let $X$ and $Y$ be random variables with mean $0$, and variances $1$ and correlation coefficient $\rho$. Show that $E[\max(X^2,Y^2)]\le 1+\sqrt{1-\rho^2}$.

My attempt:

$$E[\max(X^2,Y^2)]= E\left(\frac{X^2+Y^2-|X^2-Y^2|}{2}\right)= 1-\frac{E|X^2-Y^2|}{2}$$

which is certainly lesser than the rhs. Is there anything wrong in my reasoning?

Answer 1

Actually, the correct expression is

$$\max(X^2,Y^2)=\frac{1}{2}\left[X^2+Y^2\color{red}+|X^2-Y^2|\right]\tag{1}$$

Now simply note that

\begin{align} E|X^2-Y^2|&=E(|(X+Y)(X-Y)|) \\&=E\left[\sqrt{(X+Y)^2(X-Y)^2}\right] \\&\le \sqrt{E\left[(X+Y)^2(X-Y)^2\right]}\tag{2} \\&\le\sqrt{E\,[(X+Y)^2]E\,[(X-Y)^2]}\tag{3} \end{align}

In $(2)$, Jensen's inequality was used since $x\mapsto\sqrt x$ is a concave function for all $x\ge0$.

In $(3)$, Cauchy-Schwarz inequality was used.

So when you take expectation on both sides of $(1)$, use the above inequality and substitute the values of the expectations, you would get the desired result.