If ab + bc + ca is equal or greater than 2, what is the minimum possible value for a^2 + b^2 + c^2? This was on a noncalculator paper

1 Answer
Ananda Dasgupta
Jun 24, 2018

2

Explanation:

We start with the identity

#(a-b)^2+(b-c)^2+(c-a)^2#
#qquad = 2(a^2+b^2+c^2)-2(ab+bc+ca)#

This shows that

#2(a^2+b^2+c^2)-2(ab+bc+ca) >= 0#
(since a sum of squares of real numbers can not be negative)

Thus
#a^2+b^2+c^2 >= ab+bc+ca >= 2#

Thus the minimum possible value for #a^2+b^2+c^2 # is #2# (this minimum can be achieved if #a=b=c=sqrt{2/3}#)

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