Solve the following equation:

$$
\sqrt[4]{16+x}+\sqrt[4]{16-x}=4
$$
$$
\sqrt[4]{16+x}+\sqrt[4]{16-x}=4
$$

I. solution. Raise both sides of (1) to the fourth power and then, after rearranging the resulting equation, raise it to the fourth power again.

$$
\begin{gathered}
\sqrt{16+x}+\sqrt{16-x}=16-2 \sqrt[4]{256-x^{2}} \\
32+2 \sqrt{256-x^{2}}=256-64 \sqrt[4]{256-x^{2}}+4 \sqrt{256-x^{2}}
\end{gathered}
$$

Let's introduce the expression $\sqrt[4]{256-x^{2}}$ as a new variable, from which we get the equation

$$
y^{2}-32 y+112=0
$$

The roots of this equation are $y_{1}=4, y_{2}=28$, of which only the first can be valid, since $\sqrt[4]{256-x^{2}} \leqq 4$. From the first, we get the value $x=0$, which is indeed a root of (1).

Piriti Zsuzsa (Nagykanizsa, Landler J. Gymn., III. o. t.)

II. solution. The value of the left side of the equation at $x=0$ is 4, so this is a root of (1). We will show that if $a, b$ are any non-negative numbers, then

$$
\sqrt[4]{a}+\sqrt[4]{b} \leqq 2 \sqrt[4]{\frac{a+b}{2}}
$$

and the two sides are only equal if $a=b$. Therefore, if $x \neq 0$, the value of the left side of (1) - if it is defined - is less than 4, so (1) has only $x=0$ as a root.

First, let's prove the corresponding statement for square roots instead of (2). If $A, B$ are any non-negative numbers, then

$$
\sqrt{A}+\sqrt{B} \leqq 2 \sqrt{\frac{A+B}{2}}
$$

and the two sides are only equal if $A=B$. Since both sides are non-negative, we can take the square of both sides:

$$
A+2 \sqrt{A B}+B \leqq 2(A+B)
$$

from which we get the inequality $(\sqrt{A}-\sqrt{B})^{2} \geqq 0$. This is indeed true, and the two sides are only equal if $A=B$. Applying this statement to the numbers $A=\sqrt{a}, B=\sqrt{b}$, we get that

$$
\sqrt[4]{a}+\sqrt[4]{b} \leqq 2 \sqrt{\frac{\sqrt{a}+\sqrt{b}}{2}}
$$

Applying (3) again to the expression on the right side, we get that

$$
2 \sqrt{\frac{\sqrt{a}+\sqrt{b}}{2}} \leqq 2 \sqrt[4]{\frac{a+b}{2}}
$$

These two inequalities imply (2), and since the equality sign is only valid in them when $a=b$, we have also shown that this is the case for (2).

Remark. The inequality (2) we have proven can be written in the following form:

$$
\left(\frac{a^{1 / 4}+b^{1 / 4}}{2}\right)^{4} \leqq\left(\frac{a+b}{2}\right)
$$

The expression on the left side of the inequality is called the fourth-order power mean of $a$ and $b$. Generally, the $\alpha$-order power mean of the numbers $a_{1}$, $a_{2}, \ldots, a_{n}$ is defined as

$$
H_{\alpha}\left(a_{1}, \ldots, a_{n}\right)=\left(\frac{a_{1}^{\alpha}+\ldots+a_{n}^{\alpha}}{n}\right)^{1 / \alpha}
$$

Thus, for example, the arithmetic mean is the first-order power mean of the same numbers. The following theorem is true for power means: if $\alpha<\beta$, then the $\alpha$-order power mean is not greater than the $\beta$-order power mean, that is,

$$
H_{\alpha}\left(a_{1}, \ldots, a_{n}\right) \leqq H_{\beta}\left(a_{1} \ldots, a_{n}\right)
$$

and equality only holds if $a_{1}=a_{2}=\ldots a_{n}$. In the solution, we proved this inequality for $n=2, \alpha=1 / 4$ and $\beta=1$.