Solve $\frac{a + b}{ c }+\frac{ a + c }{b} +\frac{b + c }{a} = 4$ where $a, b$ and $c$ are integers

Question

$$\frac{a + b} { c }+\frac{ a + c }{b} + \frac{b + c }{a} = 4$$ Does this problem has solutions in integers? I tried to brute force it, but had no success for values between $-1000$ and $+1000$

a, b or c can be negative.

Answer 1

Let

$$a+b+c=S$$

$$\frac{a + b}{ c }+\frac{ a + c }{b} +\frac{b + c }{a} = 4 \implies \frac{S}{ c }-1+\frac{ S}{b}-1 +\frac{S }{a}-1 = 4 \implies \frac{S}{ c }+\frac{ S}{b}+\frac{S }{a}= 7$$

for AM-HM

$$\frac{\frac{S}{ c }+\frac{ S}{b}+\frac{S }{a}}{3}\ge \frac{3}{\frac{c}{ S }+\frac{ b}{S}+\frac{a }{S}}=3\implies {\frac{S}{ c }+\frac{ S}{b}+\frac{S }{a}} \ge 9$$

Answer 2

Assume $a, b, c > 0$, and you have: $A=(a+b+c)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge 9$ and this is known as AM-GM inequality, and the comment above implies that $7 = A$, which is absurd.