
3. Find the least positive number $x$ with the following property: if $a, b, c, d$ are arbitrary positive numbers whose product is 1 , then

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


Solution. Let $a, b, c, d$ be positive numbers whose product equals 1 . From the inequality between the arithmetic and geometric mean of the three numbers $a^{x}, b^{x}$, $c^{x}$, with arbitrary $x>0$, we obtain

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

Choosing $x=3$ gives the inequality $\frac{1}{3}\left(a^{3}+b^{3}+c^{3}\right) \geqslant 1 / d$. Similarly,

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

Adding up these four inequalities we obtain

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

so for $x=3$ the given inequality always holds.

We now show that $x=3$ is the smallest number with this property, that is, that for any positive $x1$ (depending on $xt^{3} .
$$

Thus if we can choose $t$ so that $4 t^{x}4^{\frac{1}{3-x}}
$$

which is certainly fulfilled for $t$ large enough.

Conclusion. The least number with the desired property is $x=3$.
