Let $a^2b^2+b^2c^2+a^2c^2=abc(a+b+c)$. Why does $a=b=c$?

Question

Let $a,b,c\in \mathbb{R}$ and $a,b,c \ne 0$ and $a^2b^2+b^2c^2+a^2c^2=abc(a+b+c)$.

Why does $a=b=c$?

Answer 1

$$2(a^2b^2+b^2c^2+c^2a^2)=2a^2bc+2b^2ca+2c^2ab\\\implies (a^2b^2+b^2c^2-2ab^2c)+(a^2b^2+c^2a^2-2a^2c)+(b^2c^2+c^2a^2-2abc^2)=0\\\implies (ab-bc)^2+(bc-ca)^2+(ca-ab)^2=0$$

From here you can only say $$ab=bc=ac$$

However since $a,b,c\neq0$ you get $a=b=c$

Bingo!

Answer 2

Multiply equality with 2. Then expression is equal with $(ab-bc)^2+(bc-ac)^2+(ac-ab)^2=0$

Answer 3

Let $a=cx$ and $b=cy$. The equation becomes $x^2y^2+y^2+x^2=xy(x+y+1)$, or

$$(y^2-y+1)x^2-(y^2+y)x+y^2=0$$

As a quadratic in $x$, the discriminant, which must be non-negative in order for $x$ to be real, is

$$(y^2+y)^2-4(y^2-y+1)y^2=-3y^2(y-1)^2$$

So if $y\not=0$, we must have $y=1$, which in turn implies $x=1$.