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
Aug 10, 2016

Answer:

See explanation...

Explanation:

If #a, b, c# are Real then:

#{ ((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 #ab+bc+ca# to both ends and transpose to find:

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

Aug 10, 2016

Answer:

See below

Explanation:

Given two vectors

#vec u = {a,b,c}#
#vec v = {b,c,a}#

we have

#< vec u, vec u > = norm( vec u)^2 ge < vec u, vec v >#

or equivalently

#a^2+b^2+c^2 ge ab + bc + ac#

or

#3(a^2+b^2+c^2) ge 3(ab + bc + ac)#

or

#3(a^2+b^2+c^2) -2(ab + bc + ac) ge ab + bc + ac#

but

#(a + b - c)^2 + (b + c - a)^2 + (c + a - b)^2 =3(a^2+b^2+c^2)-2(ab+bc+ac)#

so concluding

#(a + b - c)^2 + (b + c - a)^2 + (c + a - b)^2 ge ab+bc+ac#