Folkolo

Find the maximum value of the expression $ab + bc + ac + abc$, if $a + b + c = 12$ (where $a, b$, and $c$ are non-negative numbers).
The first method. $a b c \leq\left(\frac{a+b+c}{3}\right)^{3}=\left(\frac{12}{3}\right)^{3}=64$.

Moreover, $a b+b c+a c \leq a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+b c+a c)$, which means $3(a b+b c+a c) \leq(a+b+c)^{2}=$ 144.

In both cases, equality is achieved if $a=b=c=4$. Therefore, the maximum value of the given expression is $64+144: 3=112$.

The second method. Let $X=a b+b c+a c+a b c$. Then

$13+X=1+a+b+c+a b+b c+a c+a b c=(1+a)(1+b)(1+c)$

$\leq\left(\frac{3+a+b+c}{3}\right)^{3}=\left(\frac{15}{3}\right)^{3}=125$

Thus, $X \leq 112$.

Equality is achieved when $a=b=c=4$.

## Answer

112.