Thread: Inequality proof

show that

ab+ac+bc=<a^2+b^2+c^2

i've done a similar example and trying to do what i did there with this one but just not working out. below is the one i've already done

2ab=<a^2+b^2
i did like this

since

(a-b)^2=a^-2ab+b^2
thus
a^2+b^2=(a-b)^2+2ab
but
(a-b)^2>=0
then
a^2+b^2>=2ab

Originally Posted by action259

show that

ab+ac+bc=<a^2+b^2+c^2

i've done a similar example and trying to do what i did there with this one but just not working out. below is the one i've already done

2ab=<a^2+b^2
i did like this

since

(a-b)^2=a^-2ab+b^2
thus
a^2+b^2=(a-b)^2+2ab
but
(a-b)^2>=0
then
a^2+b^2>=2ab

Sounds like a good start. Try this:
(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc

If you haven't seen this before, I recommend you multiply out (a + b + c)(a + b + c) for yourself.

-Dan

ok so i get this
(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc
then
(a + b + c)^2 - 2ab - 2ac - 2bc = a^2 + b^2 + c^2
and
(a + b + c)^2>=0
so
- 2ab - 2ac - 2bc =< a^2 + b^2 + c^2
which is close to what i'm trying to show just have the multiple of 2 and negative signs so not sure this will work

Originally Posted by action259

show that

ab+ac+bc=<a^2+b^2+c^2

You can use the Cauchy-Swarthz Inequality.

Working the in vector space R^3 we define,
u=(a,b,c)
v=(b,c,a)
Then,
u*v<=||u||v||
Thus,
ab+ac+bc<=a^2+b^2+c^2