Question

If a,b,c are real numbers prove that `(a+b-c)^2`+`(b+c-a)^2`+`(c+a-b)^2)`>= ab+bc+ca ?

Answer 1

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`

Answer 2

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`