Solve the equation in the set of positive numbers:

$$
\frac{x \cdot 2014^{\frac{1}{x}}+\frac{1}{x} \cdot 2014^{x}}{2}=2014
$$
Solution. By substitution, we can verify that $x=1$ is a solution to the equation. In the following, we will show that for all other positive real numbers,

$$
\frac{x \cdot 2014^{\frac{1}{x}}+\frac{1}{x} \cdot 2014^{x}}{2}>2014
$$

Let's write the arithmetic and geometric mean inequality for the positive numbers $a=x \cdot 2014^{\frac{1}{x}}, b=\frac{1}{x} \cdot 2014^{x}$:

$$
\frac{a+b}{2} \geq \sqrt{a b}
$$

The left side is exactly the left side of the original equation. The right side can be transformed as follows:

$$
\begin{aligned}
\sqrt{a b} & =\sqrt{x \cdot 2014^{\frac{1}{x}} \cdot \frac{1}{x} \cdot 2014^{x}}=\sqrt{2014^{\frac{1}{x}} \cdot 2014^{x}}=\sqrt{2014^{\frac{1}{x}+x}}=\sqrt{2014^{\frac{x^{2}+1}{x}}}= \\
& =2014^{\frac{x^{2}+1}{2 x}}
\end{aligned}
$$

In the following, we will show that for every $x \neq 1$ positive number,

$$
2014^{\frac{x^{2}+1}{2 x}}>2014
$$

Due to the strict monotonicity of the exponential function, this is exactly

$$
\begin{gathered}
\frac{x^{2}+1}{2 x}>1, \\
\frac{x^{2}+1}{2 x}-1>0, \\
\frac{x^{2}-2 x+1}{2 x}>0 \\
\frac{(x-1)^{2}}{2 x}>0
\end{gathered}
$$

which is clearly true.

Therefore, the equation has only one positive solution, which is 1.