How to Prove a2+b2 >= 2ab and x2+y2+z2 >= 1/3 c2?

[Michael_Light]

Homework Statement

Show that a²+b² =>2ab, and hence, if x+y+z=c, show that x²+y²+z² => 1/3 c²

Homework Equations

The Attempt at a Solution

How to prove this when we only have unknowns? The only thing i can think of for the first one is a (a+b)²= a²+b² +2ab, but how to prove that a²+b² =>2ab? For the second one i have no clue...

 

[VietDao29]

Homework Statement

Show that a²+b² =>2ab

Often, we often write "greater or equal to" sign ([itex]\ge[/itex]) like this >=, to distinguish it from the "imply" sign ([itex]\Rightarrow[/itex]).

Well, you can think of (a + b)², right? So what about (a - b)²? To solve the first part of this problem, you should also note that: the square of any real number is always non-negative.

and hence, if x+y+z=c, show that x²+y²+z² => 1/3 c²

Because c = x + y + z, and you are told to prove that:

[tex]x ^ 2 + y ^ 2 + z ^ 2 \ge \frac{1}{3} c ^ 2[/tex], or written in another rather different way, you are told to prove:

[tex]x ^ 2 + y ^ 2 + z ^ 2 \ge \frac{1}{3} \left( x + y + z \right) ^ 2[/tex]

Well, I would consider expand the RHS, and notice the fact that:

x² + y² >= 2xy (as proven in the first part)

Well, let's see if you can get this problem solved. If you get stuck again, just don't hesitate to ask. :)