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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^(p-1)/pv + v^(q-1)/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^(p-1)/pv + u^(p-1)/qv
1 <= u^(p-1)/v
1 <= u^(p-1)/u^(p/q)
1 <= u^(p-1 - 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^(p-1)/pv + v^(q-1)/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^(p-1)/pv + u^(p-1)/qv      1
1 <= u^(p-1)/v                              2
1 <= u^(p-1)/u^(p/q)                    3
1 <= u^(p-1 - 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^(p-1)=q^(q-1). This does not seem to be justified since, in general, 1/(p-1) + 1/(q-1) 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^(p-1)/pv + v^(q-1)/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^(p-1)/pv + u^(p-1)/qv    (1)

Ham Fisted
Tuesday, November 18, 2003

All clear now, thanks.

james kloberdance
Tuesday, November 18, 2003

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