Find real $a$, $b$, $c$, $d$ satisfying $(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$ 
If real numbers $a$, $b$, $c$, $d$ satisfy
  $$(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$$ 
  then find $(a,b,c,d)$.

What I try: 
$$1+2a^2+2b^2+2c^2-2a-2ab-2bc=\frac{1}{4}$$
$$8a^2+8b^2+8c^2-8a-8ab-8bc+7=0$$
How do I solve it? Help me, please. 
 A: Apply CS inequality: $ LHS \ge \dfrac14\cdot ( 1- a + a - b + b - c + c )^2 = \dfrac14 = RHS $ with equality occurs when $1 - a = a - b = b - c = c$. From this you can find $a, b, c, d$. Specifically, $ b = 2c, a = 2b - c = 4c - c = 3c \implies 1 = 2a - b = 6c - 2c = 4c \implies c = \dfrac14, b = 2c = \dfrac12, a = 3c = \dfrac34$ . 
A: Let $(p,q,r,s) = (1-a,a-b,b-c,c)$, we have $p+q+r+s = 1$.
We are given
$$p^2 + q^2 + r^2 + s^2 = \frac14$$
This leads to
$$\begin{align} &\;\left(p - \frac14\right)^2 +
\left(q - \frac14\right)^2 +
\left(r - \frac14\right)^2 +
\left(s - \frac14\right)^2\\
= &\; (p^2+q^2+r^2+s^2) - \frac12(p+q+r+s) + \frac14\\
= &\; \frac14 - \frac12 + \frac14\\
= &\; 0
\end{align}
$$
As a result, $$p = q = r = s = \frac14\quad\implies\quad (a,b,c) = \left(\frac34,\frac12,\frac14\right)$$
A: You can observe this as quadratic equation on $a$ :
$$8a^2-8a(1+b) +(8b^2+8c^2  -8bc+7)=0$$ which has to have a solution so the discriminant must be nonnegative:
$$64(1+b)^2-32(8b^2+8c^2  -8bc+7)\geq 0$$
so $$6b^2-4b(1+2c)+8c^2+5\leq 0$$
Now again, the discriminat must $\geq 0$ since that inequality has a solution... 
A: Calculalte discriminant twice and the discriminat tells the value is only one. 
$2a^2-2(b+1)a+b^2+(b-c)^2+c^2+1-1/4=0 $
$D/4=(b+1)^2-2{2b^2-2bc+2c^2+3/4}\geq0$
We get,
$3b^2-2(1+2c)b+4c^2+1/2=0 $
$D/4=(1+2c)^2-3(4c^2+1/2)\geq0 $
$-8c^2+4c-1/2\geq0 $
$16(c-1/4)^2\leq0 $
$c=1/4 $
$3b^2-3b+3/4<=0 $
$(2b-1)^2\leq0 $
$b=1/2$ 
$2a^2-3a+1/4+1/16+1/16+1-1/4=0 $
$16a^2-24a+9=0 $
$(4a-3)^2=0$
$a=3/4$
