2015/022) If ab + bc + bc +cd = 16, what will at least $a^2 + b^2 + c^2 + d^2$ be?

we have

$(a-b)^2 = a^2 + b^2 - 2ab$
$(b-c)^2 = b^2 + c^2 - 2bc$
$(c-d)^2 = c^2 + d^2 - 2cd$
$(d-a) ^2 = d^2+a^2 - 2ad$

adding $(a-b)^2 + (b-c)^2 + (c-d)^2 - (d-a)^2 = 2(a^2 +b^2 +c^2 + d^2) - 2(ab+bc+cd+da)$

or $2(a^2 +b^2 +c^2 + d^2) = (a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 + 2(ab+bc+cd+da)$

or $(a^2 +b^2 +c^2 + d^2) = \dfrac{(a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2)}{2}+ (ab+bc+cd+da)$
= $\dfrac{(a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2)}{2}+ 16$

 

 clearly is lowest when a= b=c=d and value= 16

refer to https://in.answers.yahoo.com/question/index?qid=20120108032725AAASGBv