3. The sum of non-negative numbers $a, b$, and $c$ is 3. Find the maximum value of the expression $a b + b c + 2 c a$.
Answer: $\frac{9}{2}$

First solution. By the condition

$9=(a+b+c)^{2}=2(a b+b c+c a)+a^{2}+b^{2}+c^{2} \geqslant 2(a b+b c+c a)+b^{2}+2 c a \geqslant 2(a b+b c+2 c a)$.

Equality is achieved when $b=0$ and $a=c=\frac{3}{2}$. Therefore, the maximum value will be $\frac{9}{2}$.

Second solution. Notice that

$a b+b c+2 c a=(a+c)(3-a-c)+2 c a=3 a+3 c-a^{2}-c^{2}=\frac{9}{2}-\left(a-\frac{3}{2}\right)^{2}-\left(c-\frac{3}{2}\right)^{2} \leqslant \frac{9}{2}$,

and equality is achieved when $a=c=\frac{3}{2}$. Therefore, the maximum value will be $\frac{9}{2}$.