Proving $\prod_{i = 1}^n X_i \xrightarrow[n \to \infty]{(\mathbb{P})} 0 \iff \prod_{i = 1}^\infty \mathbb{E}[\sqrt{X_i}] = 0$ I am having a hard time proving the following :

Let $(X_i)_{i \geq 1}$ be a sequence of independent random variables which take their value in  $\mathbb{R}^{+*}$ and such that  $\mathbb{E}[X_i]=1$ for all  $i \geq 1$. Then prove that :
$$\prod_{i = 1}^n X_i \xrightarrow[n \to \infty]{(\mathbb{P})} 0  \iff \prod_{i = 1}^\infty \mathbb{E}[\sqrt{X_i}] = 0$$

Intuitively it seems quite natural since most of  the mass of the $X_i$ is on $[0,1]$. Taking the square-root just means we are concentrating further the mass of the $X_i$ around one. Thus it should go to zero.
Yet I don't know how to prove this result.
I tried taking the logarithm to manipulate sums yet doesn't seem to work.

Declare a positive $\varepsilon$.
For any integer $n$, you can control $E[\sqrt{(\prod_i X_i)}]$ by separating the events $\prod_i X_i < \varepsilon$ and $\prod_i X_i \geq \varepsilon$.
The first term is bounded by $\sqrt{\varepsilon}$. For the second term, use Cauchy-Schwarz : $E[1_{\prod_i X_i \geq \varepsilon } \, .\sqrt{\prod_i X_i}] \leq \sqrt{P(\prod_i X_i \geq \varepsilon)} E[\prod_i X_i] = \sqrt{P(\prod_i X_i \geq \varepsilon)}$. The hypothesis tells you this term tends to $0$ as $n$ tends to infinity.
