4

We have the following set: $\mathcal{A} = \{ \frac{m}{n} + \frac{4n}{m};\ \ m, n \in \mathbb{N} \} $

Attempting to prove that the set's minimum is 4 yields: $$\frac{m}{n}+\frac{4n}{m} = \frac{m^2 + 4 n ^2}{mn} \geq 4$$ $$m^2 + 4n^2 \geq 4mn$$ $$m^2 + 4n^2 - 4mn \geq 0$$ $$(2n-m)^2 \geq 0$$

I do not know how to proceed beyond this point, although I suspect induction may be required. Thank you for your time.

Edit: how can it be shown that the set has no upper-bound?

Asaf Karagila
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Luka Toni
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  • What's $(2n-m)^2$? – Daniel Fischer Oct 16 '13 at 12:20
  • Hint: Can $(2n-m)^2$ be negative? – Martin Oct 16 '13 at 12:29
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    Your derivation above essentially proves that the statement $\frac{m}{n} + \frac{4n}{m} \geq 4$ is equivalent to the statement $(2n-m)^2 \geq 0$. Since the latter statement is true (since a square is, after all, nonnegative), it follows that the former statement is also true. Thus you've essentially shown that $4$ is a lower bound of $\frac{m}{n} + \frac{4n}{m}$. Now show that this lower bound is actually attained, and you're done. No induction, ma! – user43208 Oct 16 '13 at 12:32
  • You messed up $;n,m;$ between your title and the body of the question... – DonAntonio Oct 16 '13 at 12:32

3 Answers3

3

Note one simple fact: For real $a,b\ge0,(\sqrt{a}-\sqrt{b})^2\ge 0\implies (a+b)\ge2\sqrt{ab}$.

Can you use that to find the lower bound?

Dust
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2

As pointed out in the comments, $(2n-m)^2$ is necessarily greater than or equal to $0,$ as the square of a real number, so it follows that $4$ is indeed a lower bound of $\mathcal A$. Now, you must find $m,n\in\Bbb N$ such that $\frac mn+\frac{4n}m=4,$ at which point you'll have shown that $4$ is in fact the minimum value of $\mathcal A$.

As for showing that $\mathcal A$ has no upper bound, consider the numbers of the form $\frac1n+\frac{4n}1.$ How large can these numbers get?

Cameron Buie
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0

To find $m, n$ such that the expression is equal to 4, set $x=\frac mn$ , so then you have to solve $x+\frac 4x =4$

Ovi
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