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My friends asked me this question about minimizing a function,

$$ E=\left( \frac{p_{2}}{p_{1}}\right)^2 + \left( \frac{p_{3}}{p_{2}}\right)^2 + ... + \left( \frac{p_{N+1}}{p_{N}}\right)^2 - N $$

where we must choose $p_{2}, p_{3}, ..., p_{N}$ so that E is minimized. I believe $p_{1}$ and $p_{N+1}$ are meant to be constants.

I thought about taking partial derivatives so we find that

$$ \frac{\partial E}{\partial p_{i}} = \frac{2p_{i}}{p_{n-1}^2} - \frac{2p_{n+1}^2}{p_{n}^3}$$

However, minimizing E in this manner would require looking at a $N \times N$ Hessian matrix, and I don't think the solution to the problem was meant to be that icky.

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  • $\begingroup$ Are you familiar with classical inequalities? $\endgroup$ – Calvin Lin Mar 11 '14 at 5:31
  • $\begingroup$ Would $p_i=0$ be a minimum? $\endgroup$ – Paul Safier May 12 '14 at 23:00
  • $\begingroup$ Nope, that would force 0's in the denominator. $\endgroup$ – DaveNine May 13 '14 at 5:19
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Ignore the $-N$ since that is a constant.

Hint: Apply QM-GM inequality. Result falls out immediately, as does the equality case.

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  • $\begingroup$ something something geometric mean inequality? Why do I feel like this is painfully obvious.. $\endgroup$ – DaveNine Mar 11 '14 at 5:39
  • $\begingroup$ @DavidSacco Quadratic Mean Geometric Mean. It states that $ \sqrt{ \sum a_i ^2 / N } \geq ( \prod a_i)^{(1/N)}$. Everything cancels out in the GM, leaving you only with $p_1$ and $p_{N+1}$. $\endgroup$ – Calvin Lin Mar 11 '14 at 5:40
  • $\begingroup$ AM-GM suffices, I suppose...+1 $\endgroup$ – Macavity Mar 11 '14 at 5:42
  • $\begingroup$ Ahhh, I see..however I guess my next question would be: how am I sure this is not just a lower bound but is, in fact, the minimum of E? $\endgroup$ – DaveNine Mar 11 '14 at 5:45
  • $\begingroup$ If the lower bound is in fact achieved, it must be the minimum. Here we can achieve it by setting $p_{k+1}=\alpha \, p_k$, where $p_1 \, \alpha^n = p_{N+1}$ $\endgroup$ – Macavity Mar 11 '14 at 6:06

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