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
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
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