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

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To prove that a² + b² ≥ 2ab, one can use the identity (a - b)² ≥ 0, which implies that a² + b² - 2ab ≥ 0. For the second part, given x + y + z = c, the goal is to show x² + y² + z² ≥ (1/3)c². This can be approached by expanding (x + y + z)² and applying the first inequality. The discussion emphasizes the importance of recognizing that the square of any real number is non-negative to establish these inequalities. Understanding these foundational concepts is crucial for solving the problems presented.
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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...
 
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Michael_Light said:

Homework Statement



Show that a2+b2 =>2ab

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

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:

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

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

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. :)
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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