X is a non-negative random variable, with $\mathbb{E}(X) < \infty $.
My goal is to show this inequality:
$$\sqrt{1+(\mathbb{E}(X))^2} \leq \mathbb{E}(\sqrt{1+X^2})$$
x² is a convex function, so with the Jensen inequality I get that:
$$\sqrt{1+(\mathbb{E}(X))^2} \leq \sqrt{1+\mathbb{E}(X^2)} = \sqrt{\mathbb{E}(1+X^2)}$$
But when I use the Jensen inequality a second time, for the concave $\sqrt{ }$ function, I get that:
$$\sqrt{1+(\mathbb{E}(X))^2} \leq \sqrt{1+\mathbb{E}(X^2)} = \sqrt{\mathbb{E}(1+X^2)} \geq \mathbb{E}(\sqrt{1+X^2}) $$
where the inequality is in the wrong direction.
Did I make a mistake? Or is more than the Jensen inequality needed to show this?
Best Answer
Try to show that the function $x\mapsto\sqrt{1+x^2}$ is convex.