If a,b,c are real numbers prove that #(a+b-c)^2#+#(b+c-a)^2#+#(c+a-b)^2)#>= ab+bc+ca ?
2 Answers
See explanation...
Explanation:
If
#{ ((a-b)^2 >= 0), ((b-c)^2 >= 0), ((c-a)^2 >= 0) :}#
So:
#0 <= 3/2((a-b)^2+(b-c)^2+(c-a)^2)#
#= 3/2(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2)#
#= 3/2(2a^2+2b^2+2c^2-2ab-2bc-2ca)#
#= 3a^2+3b^2+3c^2-3ab-3bc-3ca#
#= (a^2+b^2+c^2+2ab-2ca-2bc)+(b^2+c^2+a^2+2bc-2ab-2ca)+(c^2+a^2+b^2+2ca-2bc-2ab)-(ab+bc+ca)#
#= (a+b-c)^2+(b+c-a)^2+(c+a-b)^2-(ab+bc+ca)#
Add
#(a+b-c)^2+(b+c-a)^2+(c+a-b)^2 >= ab+bc+ca#
See below
Explanation:
Given two vectors
we have
or equivalently
or
or
but
so concluding