# Square of 3 real numbers

• Apr 27th 2010, 08:40 AM
nahduma
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?
• Apr 27th 2010, 09:16 AM
Opalg
Quote:

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
\$\displaystyle 0\leqslant (b-c)^2= b^2+c^2-2bc,\$
\$\displaystyle 0\leqslant (c-a)^2= c^2+a^2-2ca,\$
\$\displaystyle 0\leqslant (a-b)^2= a^2+b^2-2ab.\$

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

Then \$\displaystyle (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.)
• Apr 27th 2010, 09:31 AM
Wilmer
Quote:

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 (Cool)