Given integers a, b > 0, a + b = 1, show 1/(ab) + 1/(a^2 + b^2) >= 6

FriendlyAsian

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Let a, b be positive integer and a+ b= 1. Show that:
\(\displaystyle \dfrac{1}{ab} +\dfrac{1}{(a^2+b^2)} \geq 6 \)
 

Dr.Peterson

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Please show what you've tried, and where you are stuck, as always! We want to help you with what you really need, not just what we imagine might help. Even if you've tried and failed, seeing what you tried gives us a sense of where you are going wrong (if at all).
 

pka

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Let a, b be positive integer and a+ b= 1. Show that:
\(\displaystyle \dfrac{1}{ab} +\dfrac{1}{(a^2+b^2)} \geq 6 \)
If \(\displaystyle \{a,b\}\subseteq\mathbb{Z}^+\) then \(\displaystyle a+b\ne 1\).
Please correct your post.
 

topsquark

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It doesn't seem to work in any case. Using a + b = 1 try out a = 2. Your expression is negative. If \(\displaystyle a \neq b\) then try a = 2, b = 3. Your expression is 19/78.

Is there possbily a typo, beyond what pka pointed out?

-Dan
 

FriendlyAsian

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sorry guys it's not positive integers. It's just real numbers.
Besides I've figured out a solution
We have \(\displaystyle
4ab\leq (a+b)^{2}=1. \Rightarrow \dfrac{1}{ab}\geq 4.

\).
=>Left side=\(\displaystyle =\dfrac{1}{ab} +\dfrac{1}{a^2+b^2} = (\dfrac{1}{2ab}+\dfrac{1}{a^{2}+b^{2}})+\dfrac{1}{2ab}\geq \dfrac{4}{(a+b)^{2}}+\dfrac{1}{2}.4=4+2=6. \)
 

topsquark

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Are you trying to find solutions for a and b? It's not true in general. Integers are real numbers and your expression still doesn't work for a = 2, b = 1 - 2 = -1.

-Dan
 

Dr.Peterson

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The claim is true if a and b are both positive real numbers and a + b = 1. The proof given assumes this without stating it, in applying the AM-GM inequality.

Conditions are important! Time has been wasted by getting it wrong.
 
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