$ \sum_{i=1}^{10} x_i =5 $, and $ \sum_{i=1}^{10} x_i^2 =6.1 $ ; Find the largest value of these ten numbers

A General Problem

Suppose $$ \frac{\sum_{i = 1}^n x_i}{n} = a $$ and $$ \frac{\sum_{i = 1}^n x_i^2}{n} = b^2 $$ Set $y_i = x_i - a$. Then (as you can check) $$ \frac{\sum_{i = 1}^n y_i}{n} = 0 $$ So $$ b^2 = \frac{\sum_{i = 1}^n (y_i + a)^2}{n} = \frac{\sum_{i = 1}^n y_i^2 + 2ay_i + a^2}{n} = \frac{\sum_{i = 1}^n y_i^2}{n} + 2a \frac{\sum_{i = 1}^n y_i}{n} + \frac{a^2}{n} \sum_{i = 1}^n 1 $$ Setting $$ c^2 = \frac{\sum_{i = 1}^n y_i^2}{n} $$ We obtain $$ b^2 = c^2 + 2a(0) + a^2 $$ So $$ c^2 = b^2 - a^2 $$ Observe that such a sequence $y_i$ achieves its largest possible element precisely when $x_i$ achieves its largest possible element.

The upshot of all of this is that we've reduced the problem to finding the largest possible value of a sequence of numbers $y_i$ with a mean of $0$ (i.e. $\frac{\sum_{i =1}^n y_i}{n} = 0$) and variance of $c^2$ (i.e. $\frac{\sum_{i = 1}^n y_i^2}{n} = c^2$).

Intuitively, it's clear that the maximum should occur when $y_1 = \ldots = y_{n -1} = y$. Applying the mean and variance condition and performing a straightforward calculation yields $y_n = \sqrt{n - 1} c$. It remains to check that this is the best we can do.

Suppose for contradiction $y_n > \sqrt{n - 1} c$. Then $$ \sum_{i = 1}^{n - 1} y_i = -y_n $$ and $$ \sum_{i = 1}^{n - 1} y_i^2 = nc^2 - y_n^2 $$ Thus $$ \frac{\sum_{i =1}^{n - 1} y_i}{n -1} = -\frac{y_n}{n - 1} $$ and $$ \frac{\sum_{i = 1}^{n - 1} y_i^2}{n - 1} = \frac{nc^2 - y_n^2}{n - 1} $$ Set $z_i = y_i + \frac{y_n}{n - 1}$. Then $$ 0 \leq \frac{\sum_{i = 1}^{n - 1} z_i^2}{n - 1} = \frac{nc^2 - y_n^2}{n - 1} - \left(\frac{y_n}{n - 1}\right)^2 < \frac{nc^2 - (n - 1)c^2}{n -1} - \frac{(n - 1)c^2}{(n - 1)^2} = 0 $$ which is absurd. Thus $y_n = \sqrt{n - 1} c$ is the best we can do.

In terms of $x_n$, $a$, and $b$ this is $$ x_n = a + \sqrt{(n - 1)(b^2 - a^2)} $$

Your Problem

In your problem $a = .5$ and $b^2 = .61$. Plugging in, you find that $$ x_n = 2.3 $$ is the best you can do.

Just apply Cauchy-Schwarz or AM-QM inequality to the nine smaller numbers, you get (WLOG assume $x_1 \le x_2 \le \cdots \le x_{10}$)

$$ \left(\sum_{i=1}^9 x_i\right)^2 \le 9 \left( \sum_{i=1}^9 x_i^2\right)\\ \iff (5-x_{10})^2 \le 9(6.1-x_{10}^2) \\ \iff 10x_{10}^2 - 10x_{10} - 29.9 \le 0 \\ \implies x_{10} \le 2.3.\blacksquare$$