Prove That If $(a + b)^2 + (b + c)^2 + (c + d)^2 = 4(ab + bc + cd)$ Then $a=b=c=d$ If the following equation holds
$$(a + b)^2 + (b + c)^2 + (c + d)^2 = 4(ab + bc + cd)$$
Prove that $a$,$b$,$c$,$d$ are all the same.
What I did is I let $a$,$b$,$c$,$d$ all equal one number. Then I substituted and expanded. I'm sort of proud of my self (first proof I done). I'm wondering, is there another way?
(I'm teaching my self maths and I'm only a humble precalc student)

Consider the following steps
$$\begin{align}
(a + b)^2 + (b + c)^2 + (c + d)^2 &= 4(ab + bc + cd) \\
\left[ (a + b)^2-4ab \right] + \left[ (b + c)^2-4bc \right] + \left[ (c + d)^2-4cd \right] &=0 \\
\left[ a^2+b^2+2ab-4ab \right] + \left[ b^2+c^2+2bc-4bc \right] + \left[ c^2+d^2+2cd-4cd \right] &=0 \\
\left[ a^2+b^2-2ab \right] + \left[ b^2+c^2-4bc \right] + \left[ c^2+d^2-4cd \right] &=0 \\
(a-b)^2 + (b-c)^2 + (c-d)^2 &=0
\end{align}$$
and the sum of three positive numbers is zero if and only if they are all zero. So you will get
$$a=b=c=d$$
