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. :)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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