# Thread: Square of 3 real numbers

1. ## Square of 3 real numbers

Prove that, for all real numbers a,b and c,
(a+b+c)^2 <= 3(a^2 + b^2 + c^2)
Further, show that 3 is the smallest real number with this property.

I haven't been able to prove it yet - all i've done is go around in circles and play with signs. The second part of the question has me really confused. If it applies for all real numbers, how can i prove that 3 is the smallest real number with this property?

2. Originally Posted by nahduma
Prove that, for all real numbers a,b and c,
(a+b+c)^2 <= 3(a^2 + b^2 + c^2)
Further, show that 3 is the smallest real number with this property.

I haven't been able to prove it yet - all i've done is go around in circles and play with signs. The second part of the question has me really confused. If it applies for all real numbers, how can i prove that 3 is the smallest real number with this property?
Notice that
$0\leqslant (b-c)^2= b^2+c^2-2bc,$
$0\leqslant (c-a)^2= c^2+a^2-2ca,$
$0\leqslant (a-b)^2= a^2+b^2-2ab.$

Add to get $0\leqslant 2(a^2+b^2+c^2 - bc-ca-ab)$, so that $bc+ca+ab\leqslant a^2+b^2+c^2$.

Then $(a+b+c)^2 = a^2+b^2+c^2 +2(bc+ca+ab) \leqslant 3(a^2+b^2+c^2).$

To show that 3 is the smallest possible constant in that inequality, you only have to find one set of values for a,b,c for which the inequality becomes an equality. (Hint: See what happens when a=b=c.)

3. Originally Posted by nahduma
(a+b+c)^2 <= 3(a^2 + b^2 + c^2)
Well, if you equate:
(a+b+c)^2 = 3(a^2 + b^2 + c^2)
you can expand, and end up with this quadratic:
a = {b + c +- SQRT[3(2bc - b^2 - c^2)]} / 2

You can now "see" that 2bc => b^2 + c^2

Not sure if that'll help...hope so