## Task A-3.4.

Determine the smallest natural number $n$ for which there exist real numbers $x_{1}, \ldots, x_{n} \in [1,4]$ that satisfy the inequalities:

$$
\begin{aligned}
x_{1}+x_{2}+\ldots+x_{n} & \geqslant \frac{7}{3} n \\
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}} & \geqslant \frac{2}{3} n
\end{aligned}
$$
## First Solution.

Assume that for some natural number $n$ there exist real numbers $x_{1}, x_{2}, \ldots, x_{n}$ that satisfy the given conditions. Since $1 \leqslant x_{i} \leqslant 4$, the inequality holds:

$$
\left(4-x_{i}\right)\left(1-\frac{1}{x_{i}}\right) \geqslant 0
$$

where equality holds if and only if $x_{i}$ is equal to 1 or 4. Summing over all $i$, we get

$$
4 n-\left(x_{1}+x_{2}+\ldots+x_{n}\right)-4\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}\right)+n \geqslant 0
$$

On the other hand, from the conditions of the problem, it follows that

$$
4 n-\left(x_{1}+x_{2}+\ldots+x_{n}\right)-4\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}\right)+n \leqslant 4 n-\frac{7}{3} n-\frac{8}{3} n+n=0
$$

We conclude that equality must hold, which is possible only if all $x_{i}$ are in the set $\{1,4\}$.

Of the $n$ numbers $x_{1}, \ldots, x_{n}$, let $a \geqslant 0$ take the value 1, and $b \geqslant 0$ take the value 4. Then $a+b=n$, and we have

$$
\begin{array}{r}
a+4 b=x_{1}+x_{2}+\ldots+x_{n} \geqslant \frac{7}{3} n=\frac{7}{3} a+\frac{7}{3} b \\
a+\frac{1}{4} b=\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}} \geqslant \frac{2}{3} n=\frac{2}{3} a+\frac{2}{3} b
\end{array}
$$

From the first inequality, we get $4 a \geqslant 5 b$, and from the second, we get $5 a \geqslant 4 b$, from which we conclude that $4 a=5 b$. The smallest natural number $n$ for which such $a$ and $b$ exist is 9 (when $a=5, b=4$).

For $n=9$ and real numbers $x_{1}, \ldots, x_{n}$ such that five of them are equal to 1 and four are equal to 4, the conditions of the problem are satisfied, so this is indeed the smallest such natural number.