Let $Y$ be a random variable with $\mathbb{E}(Y \mid X) = X$, where $X \sim \mathrm{U}[0,1]$. Prove that $\mathbb{E}(Y^{10}) \geq \frac{1}{11}$.
I tried using Jensen's inequality and found out that $\mathbb{E}(Y^{10}) \geq \frac{1}{2^{10}}$, but it's too rough. Would thank you for any suggestions.
Based on suggestions from comments, I reproduce my Jensen's-inequality argument below:
$X\sim\mathrm{U}[0,1]$ is a r.v. uniformly distributed in $[0,1]$. By the law of total expectation, $$\mathbb{E}(Y)=\mathbb{E}(\mathbb{E}(Y\mid X))=\mathbb{E}(X)=\frac{1}{2}$$ $h(x)=x^{10}$ is convex, so by Jensen's inequality, $\mathbb{E}(h(X))\geq h(\mathbb{E}(X))$, that is $$\mathbb{E}(Y^{10})≥(\mathbb{E}(Y))^{10}=\frac{1}{210}$$
First: $$E[X^{10}]=\int_{(0,\infty)}P(X^{10}\geq t)dt=\int_{(0,1)}P(X\geq t^{1/10})dt=\int_{(0,1)}(1-t^{1/10})dt=1-\frac{10}{11}=\frac{1}{11}$$ Then by using Jensen's of conditional expectations: $$E[Y^{10}]=E[E[Y^{10}|X]]\geq E[E[Y|X]^{10}]=E[X^{10}]=\frac{1}{11}$$