## Task 23/78

For which triangles with sides $a, b, c$ does the following hold: $a^{4}+b^{4}+c^{4}=a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}$
As is well known, for $x ; y \geq 0$ the arithmetic and geometric mean inequality holds. (1) Thus,

$$
\frac{a^{4}}{2}+\frac{b^{4}}{2} \geq \sqrt{a^{4} b^{4}}=a^{2} b^{2} \quad ; \quad \frac{a^{4}}{2}+\frac{c^{4}}{2} \geq a^{2} c^{2} \quad ; \quad \frac{b^{4}}{2}+\frac{c^{4}}{2} \geq b^{2} c^{2}
$$

Adding these inequalities yields

$$
a^{4}+b^{4}+c^{4} \geq a^{2} b^{2}+a^{2} c^{2}+b^{2} c^{2}
$$

In (1), the equality sign holds if and only if $x=y$. Therefore, in (2), the equality sign holds exactly for $a=b=c$; the given inequality thus holds for all equilateral triangles and only for these.