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Is there a specific theorem to prove a^2 + b^2 > ab? Sorry this seems really stupid

 Feb 10, 2023
edited by Saphia1123  Feb 10, 2023
edited by Saphia1123  Feb 10, 2023
edited by Saphia1123  Feb 10, 2023
edited by Saphia1123  Feb 10, 2023
edited by Saphia1123  Feb 10, 2023
 #1
avatar+118616 
+3

Is there a specific theorem to prove a^2 + b^2 > ab?

 

I will assume that a and be are in the set of real numbers.

 

If and a and b are both  0 then it is not true

If one is zero but not the other than LHS is positive and RHS is 0 so it will be true

 

If a or b (but not both) is negative then the RHS is positive and the LHS is negative so it will be true

 

So I am just left to prove it is true if a and b both have the same sign. (and not 0)

 

\((a-b)^2=a^2+b^2-2ab\\so\\ LHS=a^2+b^2\\ LHS=(a-b)^2+2ab\\ LHS=ab+[(a-b)^2+ab]\\ LHS=ab + \text{a positive number}\\ LHS >ab\\ LHS>RHS\)

 Feb 10, 2023
 #2
avatar+939 
+1

Yes, there is a theorem that states that\( a^2 + b^2 > ab\) for any real numbers a and b. This theorem is known as the Cauchy-Schwarz inequality.

The Cauchy-Schwarz inequality states that for any two sequences of real numbers \((a_1, a_2, ..., a_n)\) and \((b_1, b_2, ..., b_n)\), we have:

\((a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_2^2 + ... + b_n^2) >= (a_1 b_1 + a_2 b_2 + ... + a_n b_n)^2\)

Taking \(n = 2\), we have:

\((a^2 + b^2)(b^2 + b^2) >= (ab + ab)^2\)

Expanding the right-hand side and cancelling the \(b^2\) terms, we get:

\(a^2 + b^2 > ab\)

So, for any real numbers a and b,\( a^2 + b^2 > ab\).

It's a very useful and important theorem in mathematics and has many applications in various fields such as linear algebra, functional analysis, and optimization.

 Feb 11, 2023
edited by Mathefreaker2021  Feb 11, 2023
 #3
avatar+118616 
0

Thanks Mathfreaker.    cool

 

I will point out however that It doesn't work if a and b are both 0. 

There needs to be a greater than OR equal to sign.

Melody  Feb 11, 2023

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