According to the triangle inequalities, $a+b>c, b+c>a, c+a>b$, thus

$$
\begin{array}{ccc}
c(a+b)>c^{2} & ; \quad a(b+c)>a^{2} ; & b(c+a)>b^{2} \\
c(a+b+c)>2 c^{2} \quad ; \quad a(a+b+c)>2 a^{2} & ; \quad b(c+a+b)>2 b^{2}
\end{array}
$$

By adding the three inequalities, one obtains

$$
a(a+b+c)+b(a+b+c)+c(a+b+c)=(a+b+c)^{2}>2\left(a^{2}+b^{2}+c^{2}\right)
$$

Given $a+b+c=U$ and $a^{2}+b^{2}+c^{2}=U'$, it immediately follows that $U^{2}>2 U'$ and $U^{2}: U'>2$ (since $U' \neq 0$). The lower bound of $U^{2}: U'$ (which is not attained) is therefore 2.
## Task 15/85

Let the numbers $a ; b ; c$ represent the side lengths of a triangle with perimeter $U$, and let $a^{2}$; $b^{2} ; c^{2}$ represent the side lengths of another triangle with perimeter $U^{\prime}$. Determine the lower bound of the ratio $U^{2}: U^{\prime}$!