Proving a/b+b/a >= 2 using Mathematical Proof | Homework Help

AI Thread Summary
The discussion centers on proving that a/b + b/a is greater than or equal to 2 for positive real numbers a and b. Participants suggest manipulating the expression by adding the ratios and multiplying through by ab to simplify the proof. They recognize that the inequality can be derived from the identity (a-b)² ≥ 0, which holds true for all real numbers. When a equals b, the expression equals 2, while for a not equal to b, the expression remains greater than 2. The proof ultimately relies on the non-negativity of the squared difference between a and b.
TheMathNoob
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Homework Statement


Let a,b be in the positive reals. Prove a/b+b/a is >=2

Homework Equations

The Attempt at a Solution


I have no idea. Maybe add the two ratios: (a^2+b^2)/a*b and then try to analyze separately the numerator and denominator?
 
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Those a's and b's in the denominator are pesky. Why not multiply through by ab and see what that gives you?
 
jbriggs444 said:
Those a's and b's in the denominator are pesky. Why not multiply through by ab and see what that gives you?
I feel like this is related to the law of cosines
 
TheMathNoob said:
I feel like this is related to the law of cosines
Your powers of pattern recognition are good, but there is another formula involving a2, b2 and 2ab that is simpler yet.
 
jbriggs444 said:
Your powers of pattern recognition are good, but there is another formula involving a2, b2 and 2ab that is simpler yet.
oh hahahahahah (a-b)^2>=0
when a=b (a-b)^2=0
when a!=b (a-b)^2>0
so (a-b)^2>=0
 
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