3. Find the smallest positive number $x$, for which the following holds: If $a, b, c, d$ are any positive numbers whose product is 1, then

$$
a^{x}+b^{x}+c^{x}+d^{x} \geqq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}
$$

(Pavel Novotný)
Solution. Let $a, b, c, d$ be any positive numbers whose product equals 1. According to the inequality between the arithmetic and geometric means of the triplet of numbers $a^{x}, b^{x}, c^{x}$ for any $x>0$, we have

$$
\frac{a^{x}+b^{x}+c^{x}}{3} \geqq \sqrt[3]{a^{x} b^{x} c^{x}}=\sqrt[3]{\frac{1}{d^{x}}}
$$

By choosing $x=3$, we get the inequality $\frac{1}{3}\left(a^{3}+b^{3}+c^{3}\right) \geqq 1 / d$. Similarly, we have

$$
\frac{1}{3}\left(a^{3}+b^{3}+d^{3}\right) \geqq 1 / c, \quad \frac{1}{3}\left(a^{3}+c^{3}+d^{3}\right) \geqq 1 / b, \quad \frac{1}{3}\left(b^{3}+c^{3}+d^{3}\right) \geqq 1 / a
$$

By adding the above four inequalities, we get

$$
a^{3}+b^{3}+c^{3}+d^{3} \geqq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}
$$

so for $x=3$ the inequality in the problem statement always holds.

We will show that $x=3$ is the smallest value sought, i.e., that for any positive $x<3$ (dependent on the given $x$),

$$
4 t^{x} < 4^{\frac{1}{3-x}}
$$

which is indeed satisfied for sufficiently large $t$.

Conclusion. The sought smallest positive number is $x=3$.

Note. Not only $x=3$ but also any $x \geqq 3$ (and any $x \leqq-1$) has the required property. We explain this by replacing the ones in the denominators of the fractions on the right-hand side of the considered inequality with the expression $(a b c d)^{\frac{x+1}{4}}$; we then obtain the homogeneous inequality

$$
a^{x}+b^{x}+c^{x}+d^{x} \geqq a^{\frac{x-3}{4}}(b c d)^{\frac{x+1}{4}}+b^{\frac{x-3}{4}}(a c d)^{\frac{x+1}{4}}+c^{\frac{x-3}{4}}(a b d)^{\frac{x+1}{4}}+d^{\frac{x-3}{4}}(a b c)^{\frac{x+1}{4}},
$$

which is Muirhead's inequality for the quadruples of exponents

$$
(x, 0,0,0) \text { and }\left(\frac{x+1}{4}, \frac{x+1}{4}, \frac{x+1}{4}, \frac{x-3}{4}\right),
$$

whose application in the case $x>0$ is subject to the single condition $\frac{x-3}{4} \geqq 0$ or $x \geqq 3$ (while in the case $x<0$ the only condition is $\frac{x+1}{4} \leqq 0$ or $x \leqq-1$).