For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of $\frac{a+b} {c}$ is? 
For real $a,b,c$ , if $a^2+b^2+c^2=ab+bc+ac$ ,then the value of
  $\frac{a+b} {c}$ is how much?

Ans.
What I could gather:
from the identity,
$$ a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab+bc+ca)$$
We gather that RHS=0.
$$ =>a^3+b^3+c^3=3abc$$
It would e helpful if someone could tell me how should I proceed.
 A: $a^2 + b^2 +c^2 = \frac {a^2 + b^2}{2} + \frac {b^2 + c^2}{2} + \frac {a^2 + c^2}{2}= ab + bc + ac\\
\frac {a^2 -2ab +b^2}{2} + \frac {b^2 -2bc +c^2}{2} + \frac {a^2 -2ac +c^2}{2}= 0\\
(a-b)^2 + (b-c)^2 + (c-a)^2 = 0\\
a = b = c$
$\frac {a+b}{c} = 2$
A: Hint:
$$a^2+b^2+c^2=ab+bc+ca \Leftrightarrow $$
$$\Leftrightarrow 2a^2+2b^2+2c^2=2ab+2bc+2ca \Leftrightarrow$$
$$\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0 \Leftrightarrow$$
$$\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2=0 \Leftrightarrow a=b=c$$
A: Another approach we can use is AM-GM inequality . 
$$\frac{a^2+b^2}{2} \ge \sqrt{a^2b^2}$$
$$\sum_{a,b,c}\frac{a^2+b^2}{2} \ge \sum{ab}$$
Now equality will hold only when $a=b=c$
A: For $a=b=c=0$ the needed value does not exist.
If $c\neq0$ then since 
$$0=\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2,$$
we obtain $a=b=c$ and $\frac{a+b}{c}=2$.
A: Write the relation as an equation in $c$
$$c^2 - (a + b) c + a^2 + b^2 - a b = 0$$
The discriminant is $\Delta = (a+b)^2-4(a^2 + b^2 - a b)=-3 (a - b)^2$
It is given that $c\in\mathbb{R}$ thus $\Delta$ must not be negative and the only way is to set $a=b$ and $c=\dfrac{a+b}{2}$
I like this approach not because it is mine, but because is understandable by the average student. 
It's not tricky, I mean :)
