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You can do simpler than using Chebyshev’s inequality : the distribution of $\sum_{i=1}^n X_i^2$ under $H_0$ known, it is a $\chi^2(n)$. From this you can get a threshold $t_\alpha$, using simply a chi square table: $t_\alpha = x_{n,1-\alpha}$, the quantile of level $1-\alpha$ of $\chi^2(n)$.

This is a particularly simple situation as the variance of the law under $H_0$ in known. This is almost like the kind of artificial stuff I teach to beginners, I would be pleased if you were kind enough to describe the concrete situation that lead you to this model.

You can compare the threshold you get from Chebyshev inequality: for $\sum X_i^2$ this $n\left(1+\sqrt{{2n\over\alpha}}\right)$ to the $t_\alpha$ obtained by quantiles of $\chi^2(n)$, you will see that the quantiles are much lower; this will be much more powerful than you thought.

You can do simpler than using Chebyshev’s inequality : the distribution of $\sum_{i=1}^n X_i^2$ under $H_0$ known, it is a $\chi^2(n)$. From this you can get a threshold $t_\alpha$, using simply a chi square table: $t_\alpha = x_{n,1-\alpha}$, the quantile of level $1-\alpha$ of $\chi^2(n)$.

This is a particularly simple situation as the variance of the law under $H_0$ in known. This is almost like the kind of artificial stuff I teach to beginners, I would be pleased if you were kind enough to describe the concrete situation that lead you to this model.

You can compare the threshold you get from Chebyshev inequality: for $\sum X_i^2$ this $n\left(1+\sqrt{{2n\over\alpha}}\right)$ to the $t_\alpha$ obtained by quantiles of $\chi^2(n)$, you will see that the quantiles are much lower; this will be much more powerful than you thought.

Post Scriptum To adress your original question: It is not easy to show that this test is optimal in any sense; the usual tool is Neyman-Pearson lemma but it is for tests where both $H_0$ and $H_1$ are point hypotheses, which is not the case there. Other tools are avalaible when the parameter space is $\mathbb{R}$ but here the parameters are $\mu_1, \dots, \mu_n$... So this is beyond my abilities. A family of test statistics that could be investigated is $T_\lambda = \sum_i |X_i|^\lambda$. Intuitively, I would say that the value of $\lambda$ giving the most powerful test depends on the values $(\mu_i)_{i=1,\dots,n}$. If you knew these values, the uniformely most powerful test would be given by the likelihood ratio test, this is Neyman-Pearson Lemma.

You can do simpler than using Chebyshev’s inequality : the distribution of $\sum_{i=1}^n X_i^2$ under $H_0$ known, it is a $\chi^2(n)$. From this you can get a threshold $t_\alpha$, using simply a chi square table ($t_\alpha = x_{n,1-\alpha}$, the quantile of level $1-\alpha$ of $\chi^2(n)$.

This is a particularly simple situation as the variance of the law under $H_0$ in known. This is almost like the kind of artificial stuff I teach to beginners, I would be pleased if you were kind enough to describe the concrete situation that lead you to this model.

You can compare the threshold you get from Chebyshev inequality: for $\sum X_i^2$ this $n\left(1+\sqrt{{2n\over\alpha}}\right)$ to the $t_\alpha$ obtained by quantiles of $\chi^2(n)$, you will see that the quantiles are much lower; this will be much more powerful than you thought.

You can do simpler than using Chebyshev’s inequality : the distribution of $\sum_{i=1}^n X_i^2$ under $H_0$ known, it is a $\chi^2(n)$. From this you can get a threshold $t_\alpha$, using simply a chi square table: $t_\alpha = x_{n,1-\alpha}$, the quantile of level $1-\alpha$ of $\chi^2(n)$.

This is a particularly simple situation as the variance of the law under $H_0$ in known. This is almost like the kind of artificial stuff I teach to beginners, I would be pleased if you were kind enough to describe the concrete situation that lead you to this model.

You can compare the threshold you get from Chebyshev inequality: for $\sum X_i^2$ this $n\left(1+\sqrt{{2n\over\alpha}}\right)$ to the $t_\alpha$ obtained by quantiles of $\chi^2(n)$, you will see that the quantiles are much lower; this will be much more powerful than you thought.

You can do simpler than using Chebyshev’s inequality : the distribution of $\sum_{i=1}^n X_i^2$ under $H_0$ known, it is a $\chi^2(n)$. From this you can get a threshold $t_\alpha$, using simply a chi square table.

This is a particularly simple situation as the variance of the law under $H_0$ in known. This is almost like the kind of artificial stuff I teach to beginners, I would be pleased if you were kind enough to describe the concrete situation that lead you to this model.

You can do simpler than using Chebyshev’s inequality : the distribution of $\sum_{i=1}^n X_i^2$ under $H_0$ known, it is a $\chi^2(n)$. From this you can get a threshold $t_\alpha$, using simply a chi square table ($t_\alpha = x_{n,1-\alpha}$, the quantile of level $1-\alpha$ of $\chi^2(n)$.

This is a particularly simple situation as the variance of the law under $H_0$ in known. This is almost like the kind of artificial stuff I teach to beginners, I would be pleased if you were kind enough to describe the concrete situation that lead you to this model.

You can compare the threshold you get from Chebyshev inequality: for $\sum X_i^2$ this $n\left(1+\sqrt{{2n\over\alpha}}\right)$ to the $t_\alpha$ obtained by quantiles of $\chi^2(n)$, you will see that the quantiles are much lower; this will be much more powerful than you thought.