3.1. Find all real values of $\alpha$ for which the following holds: if $a, b, c$ are the sides of a triangle, then

$$
a^{2}+b^{2}+c^{2} \leq \alpha(a b+a c+b c)
$$
## 3.1. In a triangle, the following holds

$$
\begin{gathered}
a+b-c>0, a-b+c>0, -a+b+c>0 \\
(a+b-c)(a-b+c)+(a+b-c)(-a+b+c)+ \\
+(a-b+c)(-a+b+c)= \\
=2(a b+a c+b c)-(a^{2}+b^{2}+c^{2})>0
\end{gathered}
$$

i.e., the desired inequality holds for $\alpha \geq 2$. Let's take $a=b=1$ (so $0<c<2$). We obtain the inequality $c^{2}-2 \alpha c-\alpha+2 \leq 0$, which for $\alpha<2$ will not hold if we take a sufficiently small $c$.