Undergraduate math problem
Here is a seemingly harmless problem I encountered years ago while earning my undergraduate math degree which I have never been able to solve: if P and Q are positive numbers and (1/q + 1/p) = 1; prove that if u is greater than or equal to zero and if v is greater than or equal to zero then, u*v is less than or equal to (u^p)/p + (v^q)/q.
James kloberdance
Saturday, November 15, 2003
It's just calculus isn't it?
Last inequality is always true when u or v equal zero, so consider u>0 and v>0.
Divide by uv.
1 <= u^(p1)/pv + v^(q1)/qu
Now find the minimum value the right hand side can take. Differentiating w.r.t. either u or v and setting the differential equal to zero gives you u^p = v^q. This is the minimum because the right hand side goes to +inf as u approaces zero or as u approaches infinity, and likewise with v. So we need to show
1 <= u^(p1)/pv + u^(p1)/qv
1 <= u^(p1)/v
1 <= u^(p1)/u^(p/q)
1 <= u^(p1  p/q)
1 <= u^(p(1  1/p  1/q))
1 <= u^0
which is true. QED
Ham Fisted
Tuesday, November 18, 2003
Thanks for looking at this problem. I have a few quetions that I have to ask before I can accept your proof.
Undergraduate math problem
Here is a seemingly harmless problem I encountered years ago while earning my undergraduate math degree which I have never been able to solve: if P and Q are positive numbers and (1/q + 1/p) = 1; prove that if u is greater than or equal to zero and if v is greater than or equal to zero then, u*v is less than or equal to (u^p)/p + (v^q)/q.
James kloberdance
Saturday, November 15, 2003
It's just calculus isn't it?
Last inequality is always true when u or v equal zero, so consider u>0 and v>0.
Divide by uv.
1 <= u^(p1)/pv + v^(q1)/qu
Now find the minimum value the right hand side can take. Differentiating w.r.t. either u or v and setting the differential equal to zero gives you u^p = v^q. This is the minimum because the right hand side goes to +inf as u approaces zero or as u approaches infinity, and likewise with v. So we need to show
1 <= u^(p1)/pv + u^(p1)/qv 1
1 <= u^(p1)/v 2
1 <= u^(p1)/u^(p/q) 3
1 <= u^(p1  p/q) 4
1 <= u^(p(1  1/p  1/q)) 5
1 <= u^0 6
which is true. QED
Question number 1: It seems as if you were right to conclude that u^p = v^q when you are minimizing the above function, however, later you use this fact to say u^(p1)=q^(q1). This does not seem to be justified since, in general, 1/(p1) + 1/(q1) does not equal 1.
Question number 2: In line two you are saying that v = u^(p/q)...this does not seem to be correct.
james kloberdance
Tuesday, November 18, 2003
I don't see where I used the eqn you said in question 1, could you point out the step?
as for question 2, if u^p = v^q, and all variables are positive, can't you raise both sides to 1/q and get u^(p/q) = v? I may be rusty on the algebra.
Ham Fisted
Tuesday, November 18, 2003
oh I see where you are getting question 1 from. The answer is, to arrive at eqn 1, I took
1 <= u^(p1)/pv + v^(q1)/qu (0)
1 <= u^p/pvu + v^q/quv (0.1)
and substituted u^p for v^q.
1 <= u^p/pvu + u^p/quv (0.2)
1 <= u^(p1)/pv + u^(p1)/qv (1)
Ham Fisted
Tuesday, November 18, 2003
All clear now, thanks.
james kloberdance
Tuesday, November 18, 2003
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