For each value for a and b, being real numbers, a:2+b:2 >= ab

Yaretzi Mcconnell 2022-11-15 Answered
For each value for a and b, being real numbers, a 2 + b 2 a b
Should I solve this by replacing all possible real number formulas in these to be able to prove this? As an example for odd numbers, by replacing 2 x + 1, etc...?
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Answers (2)

Calvin Maddox
Answered 2022-11-16 Author has 15 answers
Step 1
Try to write it as a 2 + b 2 a b = 2 a 2 + 2 b 2 2 a b 2 = a 2 + b 2 + ( a b ) 2 2 .
Step 2
Now all three terms are 0, so you get a 2 + b 2 a b.
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Kareem Mejia
Answered 2022-11-17 Author has 9 answers
Step 1
Alternative:
Either ( a × b ) is negative, or it isn't.
Case 1:   ( a × b )   is negative _
Then, a 2 + b 2 0 > ( a × b ) .
Step 2
Case 2:   ( a × b )   is not negative _
Then 2 ( a × b ) ( a × b ) .
Further, since ( a b ) 2 0 you have that
a 2 + b 2 2 ( a × b ) ( a × b ) .
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