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Good morning everyone!

By the arithmetic-geometric mean inequality, we all know that a suitable lower bound for the quantity

$$\frac{a}{b} + \frac{b}{a}$$

is $2$.

Now my question is: Will this quantity ever be bounded from above, if we allow $a$ and $b$ to be in $\mathbb{R}$?

If the answer is NO, under what conditions on $a$ and $b$ can we ensure the existence of an upper bound?

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If $b=1$ it's $a+1/a$ which becomes infinite with $a$. – coffeemath Mar 27 '13 at 13:49
what does the «ever» mean in the title? – Mariano Suárez-Alvarez Mar 27 '13 at 13:50
<ever> in the sense of <being possible>, thanks for clarifying that particular detail Mariano... =) – user11235813 Mar 27 '13 at 14:17
Why not ask the much more clear «is $\frac ab+\frac ba$ bounded above?»? – Mariano Suárez-Alvarez Mar 27 '13 at 16:01
Thank you for your clarificatory question, @Mariano. My answer would be: Because the answer to "your" last question would depend on whether there is indeed a specific relationship between $a$ and $b$ (as AndyBrandi has correctly pointed out). I was trying to discount that possibility and ask whether the sum remains bounded under "other conditions" (i.e., when there is no relationship [either implicit or explicit] between $a$ and $b$ -- in other words, when $a$ is "independent" of $b$). – user11235813 Mar 28 '13 at 15:03
up vote 6 down vote accepted

We have $$ \lim_{a\rightarrow 0}\; \left|\frac{a}{b} + \frac{b}{a}\right| = \infty$$ and $$ \lim_{a\rightarrow \infty}\; \left|\frac{a}{b} + \frac{b}{a}\right| = \infty$$ regardless of your choice of $b$ (and vice-versa), so you won't be able to give an upper bound unless there is some relationship between $a$ and $b$.

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but even so we would have something like $k + 1/k$ which goes to infinite. We need to bound $a$ and $b$, and have them $\ge \varepsilon \gt 0$. – mau Mar 27 '13 at 14:15
Thanks @AndyBrandi! Does that mean that, if there is no relationship between $a$ and $b$, then there is no (possible) upper bound? – user11235813 Mar 27 '13 at 14:18
@mau, indeed - for the problem I'm considering, I already have both $a \geq \epsilon_1 > 0$ and $b \geq \epsilon_2 > 0$ (and then, consequential bounds for $a$ and $b$ under particular cases). My problem is that, I am not sure if these restrictions would ensure that $\frac{a}{b} + \frac{b}{a}$ does not blow-up, if you know what I mean... =) – user11235813 Mar 27 '13 at 14:21
if you don't have any relation between $\varepsilon_1$ and $\varepsilon_2$, then their ratio may go to infinite anyway. – mau Mar 27 '13 at 14:28
Sorry, I just expressed myself poorly. My point was that there do exist unbounded subsets $D\subset\mathbb{R}^2$ such that $$\sup_{(a,b)\in D} \left| \frac{a}{b}+\frac{b}{a} \right|<\infty.$$ $\left\{ (a,b)\in\mathbb{R}^2: a=b \right\}-(0,0)$ obviously is such a set. – Andy Brandi Mar 27 '13 at 15:11

Try $a = 1$ and $b \to \infty$. I would guess that it is bounded, iff $a/b$ and $b/a$ are bounded. So you don't win anything...

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Thanks @gerw! If $\frac{a}{b}$ is bounded and $b < a$, then automatically $\frac{b}{a}$ will also be bounded. So yes, I agree, I still don't "win anything" at this point... =) – user11235813 Mar 27 '13 at 14:24

As most of the other answerers have noted, there is no global upper bound on $\frac xy + \frac yx$ because as either of them approaches 0, the term tends toward infinity.

However, your second part of the question has many different answers. For example, if you constrain $1 < x < 10$ and $1 < y < 10$, and then the upper bound will be $10.1$. In fact, you can even set a constraint $\frac xy + \frac yx < 3$ and the upper bound of that set will be $3$.

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Just to clarify, if $$\frac{x}{y} + \frac{y}{x} < 3$$ then $1 < x < 3$ and $1 < y < 3$? – user11235813 Mar 27 '13 at 15:19
Not quite. $x$ and $y$ can be as large as you like, as long as their ratio is within a certain bound. – Joe Z. Mar 27 '13 at 15:23
Okay, thanks Joe! =) – user11235813 Mar 27 '13 at 15:37
For example, if $x = 1000000$ and $y = 1000000$, you still have that $\frac xy + \frac yx = 2$. – Joe Z. Mar 27 '13 at 15:38
Indeed @JoeZeng... =) – user11235813 Mar 27 '13 at 15:39

$(\dfrac{\sqrt a}{\sqrt b} - \dfrac{ \sqrt b}{\sqrt a})^2 \ge 0$

$\dfrac{a}{b} + \dfrac{b}{a} \ge 2$

Now you know it works for $a$ and $b$ being reals, too.

When you have to get the upper bound of the expression. Let one of the values to be maximum and other to minimum.

In this case:

$$\lim_{a\to \infty, b \to 0}\dfrac{a}{b} + \dfrac{b}{a} = \infty$$. Therefore upperbound of the expression is tending to $\infty$

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Thanks @Inceptio. Can you clarify why $${\left(\frac{\sqrt{a}}{\sqrt{b}} - \frac{\sqrt{b}}{\sqrt{a}}\right)}^2 \geq 0$$ is a viable starting point? I know it does give the correct lower bound, following AM-GM, but I would be interested to know if we will be able to get any upper bounds using exactly the same starting point... =) – user11235813 Mar 27 '13 at 14:27
I have started that way to show that AM-GM works for reals. And there is no need of using the same start, as the expression itself clearly states about bound being tending to $\infty$. – Inceptio Mar 27 '13 at 14:30
Okay, thanks again @Inceptio... =) – user11235813 Mar 27 '13 at 15:18

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