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Prove that $$\mathbb{E}(\sqrt{X})\leq\sqrt{\mathbb{E}Y}$$ where random variables $X,Y>0$ and $\mathbb{E}\left[\frac{X}{Y}\right]\leq1.$

My attempt: This looks very much like Jensen's inequality.

According to Jensen, $\mathbb{E}(\sqrt{X})\leq\sqrt{\mathbb{E}X},$ then the desired inequality is true if we can prove $\mathbb{E}X\leq\mathbb{E}Y.$ But this seems impossible to prove because $X$ and $Y$ are not independent.

Any help will be appreciated!

lovelesswang
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1 Answers1

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The Cauchy-Schwarz inequality gives: $$\mathbb{E}[\sqrt{X}]=\mathbb{E}\left[\sqrt{Y}\cdot\sqrt{\frac{X}{Y}}\right]\leq\sqrt{\mathbb{E}[Y]\cdot\mathbb{E}\left[\frac{X}{Y}\right]}\leq\sqrt{\mathbb{E}[Y]}.$$

Jack D'Aurizio
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