Problem 9.5. For real non-zero numbers $a, b, c$ it is known that

$$
a^{2}-b c=b^{2}-a c=c^{2}-a b
$$

(a) (1 point) What positive values can the expression

$$
\frac{a}{b+c}+\frac{2 b}{a+c}+\frac{4 c}{a+b} ?
$$

take? List all possible options. If the expression cannot take positive values, write 0 as the answer.

(b) (3 points) What negative values can the expression

$$
\frac{a}{b+c}+\frac{2 b}{a+c}+\frac{4 c}{a+b} ?
$$

take? List all possible options. If the expression cannot take negative values, write 0 as the answer.
# Answer:

(a) (1 point) $7 / 2$.

(b) (3 points) -7.

Solution. First case: all numbers $a, b, c$ are equal to each other.

Then

$$
\frac{a}{b+c}+\frac{2 b}{a+c}+\frac{4 c}{a+b}=\frac{1}{2}+\frac{2}{2}+\frac{4}{2}=\frac{7}{2}
$$

Second case: among $a, b, c$ there are different numbers.

Without loss of generality, $a \neq b$, then

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

This same result will be obtained if $a \neq c$ or $b \neq c$. That is, if among $a, b, c$ there are two different numbers, then the sum of all three numbers is zero. We get that

$$
\frac{a}{b+c}+\frac{2 b}{a+c}+\frac{4 c}{a+b}=\frac{a}{-a}+\frac{2 b}{-b}+\frac{4 c}{-c}=-7
$$