Prove Inequality: 1/p + 1/q=1, u,v >= 0

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SUMMARY

The discussion centers on proving the inequality \( uv \leq \frac{u^p}{p} + \frac{v^q}{q} \) given the condition \( \frac{1}{p} + \frac{1}{q} = 1 \) for positive real numbers \( p \) and \( q \). Participants suggest using the weighted AM-GM inequality as a potential approach to the proof. The rearrangement of the initial condition to \( p + q = pq \) is also highlighted as a crucial step in the solution process. The conversation emphasizes the need for clarity on which inequalities are permissible for the proof.

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  • Understanding of the AM-GM inequality
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  • Knowledge of weighted means and their properties
  • Basic concepts of real analysis involving inequalities
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  • Study the weighted AM-GM inequality in detail
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  • Review the implications of the condition \( \frac{1}{p} + \frac{1}{q} = 1 \) in various contexts
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Students and educators in mathematics, particularly those focusing on real analysis and inequality proofs, as well as anyone interested in advanced algebraic techniques.

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Homework Statement


Let p and q be positive real numbers such that

1/p + 1/q=1

Prove that if u\geq 0 and v \geq 0, then

uv \leq \frac{u^p}{p}+\frac{v^q}{q}


Homework Equations





The Attempt at a Solution


I am really stumped. Is there like a famous inequality that I need to use here that I am forgetting?
This vaguely reminds me of the AM-GM inequality...
 
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It certainly works when u=v=1...
 
I think I got it, but it was tricky.

First, try to prove that \frac{1}{p}+\frac{1}{q}=1 can be rearranged to show that p+q=pq

Then start to work on rearranging the second inequality, in order to remove the fractions.
 
CrazyIvan said:
I think I got it, but it was tricky.

First, try to prove that \frac{1}{p}+\frac{1}{q}=1 can be rearranged to show that p+q=pq

Well, that comes from just multiplying the first equation by pq.



CrazyIvan said:
Then start to work on rearranging the second inequality, in order to remove the fractions.

So, I have been trying this and not getting anywhere. I tried plugging in p q = p+q. Can you be more specific?
 

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