$$
\begin{aligned}
& K=a^{4}\left(b^{2} c+b c^{2}\right)+a^{3}\left(b^{3} c+b c^{3}\right)+a^{2}\left(b^{3} c^{2}+b^{2} c^{3}+b^{2} c+b c^{2}\right)+a\left(b^{3} c+b c^{3}\right)+ \\
&+\left(b^{3} c^{2}+b^{2} c^{3}\right) \geq 12 a^{2} b^{2} c^{2} .
\end{aligned}
$$

When does equality hold?
I. solution. On the left side, after performing the multiplication, there is a sum of 12 terms. Their geometric mean is

$$
\sqrt[12]{a^{24} b^{24} c^{24}}=a^{2} b^{2} c^{2}
$$

which is one-twelfth of the right side. According to this, we are faced with the inequality between the arithmetic and geometric means of the 12 terms on the left side, multiplied by 12.

Equality holds if and only if all 12 terms are equal. The condition for this is that the equality of any two-term factor implies $b=c$, and with this, the four-term factor's 2-2 terms are also equal. From the same, $b^{2} c^{3}=b^{2} c$ implies $c^{2}=1$, and thus $c=1=b$. Finally, from the equality of the 2nd and 4th bracketed expressions, $a^{2}=1, a=1$. Indeed, when $a=b=c=1$, the value of each of the 12 terms is 1, and the right side is 12.

György Farkas (Budapest, Landler J. Gép- és Hír. Ip. T., II. o. t.)

Essentially, the other solution types also use the inequality between the arithmetic and geometric means of positive numbers, applying it to fewer terms at once, in different groupings.

II. solution. On the left side of (1), there are six 7th and 5th degree terms. Pairing them appropriately - i.e., the terms from the front and back that are the same number - 

$$
\begin{gathered}
K=\left(a^{4} b^{2} c+b^{2} c^{3}\right)+\left(a^{4} b c^{2}+b^{3} c^{2}\right)+\left(a^{3} b^{3} c+a b c^{3}\right)+\left(a^{3} b c^{3}+a b^{3} c\right)+ \\
+\left(a^{2} b^{3} c^{2}+a^{2} b c^{2}\right)+\left(a^{2} b^{2} c^{3}+a^{2} b^{2} c\right)
\end{gathered}
$$

the geometric mean of each pair is $a^{2} b^{2} c^{2}$, and each bracket is $\geq 2 a^{2} b^{2} c^{2}$. - Equality holds in (1) if and only if it holds in each of the rearranged brackets. From the 6th and 5th brackets, $c^{2}=1, b^{2}=1-$ from which $c=b=1-$, and based on these, from the fourth, $a^{2}=1, a=1$.

Gábor Eszes (Budapest, Berzsenyi D. g. III. o. t.)

Remark. The following rearrangement also leads to the same result:

$$
\begin{aligned}
K=\left(a^{4} b c^{2}+\right. & \left.b^{3} c^{2}\right)+\left(a^{4} b^{2} c+b^{2} c^{3}\right)+\left(a^{2} b c+b c\right) \cdot\left(a b^{2}+a c^{2}\right)+ \\
& +\left(a^{2} b^{3} c^{2}+a^{2} b c^{2}\right)+\left(a^{2} b^{2} c^{3}+a^{2} b^{2} c\right)
\end{aligned}
$$