If both roots of the Quadratic Equation are similar then prove that If both roots of the equation $(a-b)x^2+(b-c)x+(c-a)=0$ are equal, prove that $2a=b+c$.

Things should be known:
  
  
*
  
*Roots of a Quadratic Equations can be identified by:
The roots can be figured out by: 
  $$\frac{-b \pm \sqrt{d}}{2a},$$ 
  where
  $$d=b^2-4ac.$$
  
*When the equation has equal roots, then $d=b^2-4ac=0$.
  
*That means $d=(b-c)^2-4(a-b)(c-a)=0$

 A: The expression for $d$ factors as $(2a-b-c)^2$, so we must have $2a-b-c=0$. 
The easiest way to see this is to let $x=a-b,y=c-a$ and note that $b-c=-x-y$: $d=(-x-y)^2-4xy=x^2+y^2+2xy-4xy=(x-y)^2$.
Alternative solution: Note that $(a-b)+(b-c)+(c-a)=0$, i.e. $x=1$ is a root of the equation. If both roots are equal, it means that the other root is also $1$, showing that $(a-b)x^2+(b-c)x+(c-a)=(a-b)(x-1)^2$.
Equating the constant term, we find: $c-a=a-b$, which means $2a=b+c$. (This can be done with the coefficient of $x$ too).
A: As the two roots are equal the discriminant must be equal to $0$.
$$(b-c)^2-4(a-b)(c-a)=(a-b+c-a)^2-4(a-b)(c-a)=\{a-b-(c-a)\}^2=(2a-b-c)^2=0 \iff 2a-b-c=0$$

Alternatively, solving for $x,$ we get $$x=\frac{-(b-c)\pm\sqrt{(b-c)^2-4(a-b)(c-a)}}{2(a-b)}=\frac{c-b\pm(2a-b-c)}{2(a-b)}=\frac{c-a}{a-b}, 1$$ as $a-b\ne 0$ as $a=b$ would make the equation linear. 
So, $$\frac{c-a}{a-b}=1\implies c-a=a-b\implies 2a=b+c$$
A: Edit: $(b-c)^2=4(a-b)(c-a)$, so $a-b$ and $c-a$ are either both nonnegative or both nonpositive. In the former case, put $x=a-b$ and $y=c-a$; in the latter case, put $x=-(a-b)$ and $y=-(c-a)$. Now the A.M. $\ge$ G.M. inequality says that
$$
\frac{x+y}2\ge\sqrt{xy},\ \text{ i.e. } (x+y)^2\ge4xy
$$
for $x,y\ge0$, and equality holds if and only if $x=y$. 
