<s>[INST] 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. 
 [/INST] 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$ . 
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