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Prove that, $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}\ge 4$ where we do not use AM-GM inequality on the given statement to prove it. Typically, I am actually looking for a little advanced and elegant solution.

EDIT: $a,b,c,d>0$

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  • $\begingroup$ Lagrange multipliers? It seems obvious that at the minimum point of some well-chosen open set, you'll have some symmetry on the minimizer, which will in turn imply that the minimum is attained at $a=b=c=d$, which gives you $4$ as the minimum. $\endgroup$ Feb 23, 2014 at 5:48
  • $\begingroup$ I do not have any problem with that until I can understand that. $\endgroup$
    – Hawk
    Feb 23, 2014 at 5:49
  • $\begingroup$ Do you assume $a,b,c,d > 0$? or just $\neq 0$? $\endgroup$ Feb 23, 2014 at 5:51
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    $\begingroup$ Yes, they are obviously positive else, we cannot use AM-GM inequality. $\endgroup$
    – Hawk
    Feb 23, 2014 at 5:52

5 Answers 5

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Not exactly AM-GM, but Rearrangement Inequality immediately gives

$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \ge \frac{a}{a}+\frac{b}{b}+\frac{c}{c}+\frac{d}{d} = 4$$

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  • $\begingroup$ Yes, it is correct and quite elegant. $\endgroup$
    – Hawk
    Feb 23, 2014 at 7:20
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For $N > 0$, consider the set $$ D_N = \left\{ (a,b,c,d) \in \mathbb R^4 \, | \, \frac 1N \le a,b,c,d \le N \right\}. $$ This set is compact, and the function $f(a,b,c,d) \overset{def}= \frac ab + \frac bc + \frac cd + \frac da$ is continuous on $D_N$, hence we have the existence of a minimum. Without loss of generality, we can assume that there is a minimizer in the interior of $D_N$, since if it is on the boundary for some $N$, it is in the interior of a $D_N$ for $N$ slightly bigger. (Note that when any component approaches $0$ or $\infty$, $f \to \infty$, hence the minimum is not there, which allows us to consider the interior only and dismiss boundaries.)

Computing the gradient, one gets $$ \nabla f(a,b,c,d) = \left( \frac 1b - \frac d{a^2}, \frac 1c - \frac a{b^2}, \frac 1d - \frac b{c^2}, \frac 1a - \frac c{d^2}\right) $$ and since our minimizer will be in the interior, there exists a minimum (A,B,C,D) such that $$ A^2 = BD, B^2 = CA, C^2 = DB, D^2 = AC \implies A=B=C=D \implies f(A,B,C,D) = 4. $$ Since $4$ is the minimum value for all $N$, it is the minimal value.

Hope that helps,

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  • $\begingroup$ (+1)Thank you, but don't you think...it is a liitle extensive solution compared to the level of the problem? $\endgroup$
    – Hawk
    Feb 23, 2014 at 6:02
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    $\begingroup$ @Hawk : I gave it a thought and actually I'm just used to saying "Lagrange multipliers" when I minimize stuff... but I only used the fact that at the minimum in an interior, the gradient is zero... there's nothing really special about that. $\endgroup$ Feb 23, 2014 at 6:04
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    $\begingroup$ Thank you for the approach $\endgroup$
    – Hawk
    Feb 23, 2014 at 6:05
  • $\begingroup$ A very nice solution indeed,but as the OP says,it might be a bit difficult for him/her-I don't think they have any experience yet in the subtleties of theoretical calculus. $\endgroup$ Feb 23, 2014 at 6:06
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    $\begingroup$ Wait, if you make $N$ slightly bigger, that minimum point is now in the interior, but it might not be a minimum in the new region. $\endgroup$ Feb 23, 2014 at 6:09
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Let's prove that $\frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} + \frac{d}{d'} \geq 4$ if $(a', b', c', d')$ is the tuple $(a, b, c, d)$ shuffled in an arbitrary order. Your inequality will follow as a special case when $(a', b', c', d')=(b, c, d, a)$.

When the problem is formulated like this, we can safely assume that $a \leq b \leq c \leq d$. Let's do that to make things easier.

If $(a', b', c', d') \neq (a, b, c, d)$, then there's a pair of adjacent elements in the tuple $(a', b', c', d')$ that goes in decreasing order. For instance, let's assume that $a' > b'$. Form a new tuple $(a'', b'', c'', d'') = (b', a', c', d')$. We have swapped positions of the two elements so that now they go in the ascending order. Observe that since $a' > b'$ and $a < b$, we have $$ \frac{a}{a''} + \frac{b}{b''} = \frac{a}{b'} + \frac{b}{a'} \leq \frac{a}{a'} + \frac{b}{b'}, $$ therefore $$ \frac{a}{a''} + \frac{b}{b''} + \frac{c}{c''} + \frac{d}{d''} \leq \frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} + \frac{d}{d'}. $$

Using this observation, we can do the following. We can take the tuple $(a', b', c', d')$ and bubble sort it. Each time when we swap places of two adjacent elements that go in descending order, the value $\frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} + \frac{d}{d'}$ decreases, as we have just shown. When we are done with the bubble sort, we will have the tuple that goes in ascending order, i.e. $(a', b', c', d') = (a, b, c, d)$. For this tuple we have $\frac{a}{a} + \frac{b}{b} + \frac{c}{c} + \frac{d}{d} = 4$. Since the value was only decreasing during the bubble sort, the original value was greater or equal than the final one. Therefore $\frac{a}{a'} + \frac{b}{b'} + \frac{c}{c'} + \frac{d}{d'} \geq 4$, for an arbitrary arrangement $(a', b', c', d')$, qed.

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  • $\begingroup$ If you use the rearrangement inequality instead of reproving it, your solution becomes much simpler. $\endgroup$
    – vadim123
    Feb 23, 2014 at 6:24
  • $\begingroup$ @vadim123 You are right. I wanted to do just that initially, but then decided otherwise. A questionable decision, I agree. $\endgroup$
    – Dan Shved
    Feb 23, 2014 at 6:28
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$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right)=$$

$$=\left[\sqrt{\frac{a}{b}+\frac{c}{d}}-\sqrt{\frac{b}{c}+\frac{d}{a}}\right]^2+2\sqrt{\left(\frac{a}{b}+\frac{c}{d}\right)\cdot\left(\frac{b}{c}+\frac{d}{a}\right)}=$$

$$=\left[\sqrt{\frac{a}{b}+\frac{c}{d}}-\sqrt{\frac{b}{c}+\frac{d}{a}}\right]^2+2\sqrt{\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{d}+\frac{d}{b}\right)}=$$

$$=\left[\sqrt{\frac{a}{b}+\frac{c}{d}}-\sqrt{\frac{b}{c}+\frac{d}{a}}\right]^2+2\sqrt{4+\left(\sqrt{\frac{a}{c}}-\sqrt{\frac{c}{a}}\right)^2+\left(\sqrt{\frac{b}{d}}-\sqrt{\frac{d}{b}}\right)^2}\ge$$

$$\ge 2\sqrt{4}=4.$$

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    $\begingroup$ You essentially iterated AM-GM. $\frac{x+y}2 \ge \sqrt{xy}$ is AM-GM. $\endgroup$ Feb 24, 2014 at 3:27
  • $\begingroup$ I show the inequality for two numbers only directly without using AM-GM. If I used, say derivatives to show it, would it be less AM-GM? And the whole purpose is to rearrange terms so that I can apply it three times. Nothing in this solution is used besides the fact that a square is non-negative. $\endgroup$
    – Vadim
    Feb 24, 2014 at 16:34
  • $\begingroup$ Showing AM-GM for two numbers doesn't make it less AM-GM... It so happens that in the case of two numbers AM-GM is equivalent to a square being positive. I understand that you avoided explicitly quoting it but still. I mean, let OP decide. $\endgroup$ Feb 24, 2014 at 18:08
  • $\begingroup$ I can assure you that in math you cannot avoid inequalities that can be considered as AM-GM. $1\ge1$ is also a AM-GM inequality for one variable. :) And a square is positive not because of the AM-GM. Moreover, a way to prove AM-GM is actually based on this fact. Anyway, I will rewrite it to one-line avoiding using AM-GM in any form. $\endgroup$
    – Vadim
    Feb 24, 2014 at 21:01
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    $\begingroup$ I don't, just clarifying. Anyway, I rewrote it to not include any inequalities up to the last step, where I simply use that $x^2\ge 0$. $\endgroup$
    – Vadim
    Feb 25, 2014 at 2:25
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I don't know whether the following argument will be considered as AM-GM (basically, this is the way one proves AM-GM for 4 terms)

Take $a=x^2, b=y^2, c=z^2, d=t^2$. Then we have $\frac{a}{b}+\frac{d}{a}=\frac{x^2}{y^2}+\frac{t^2}{x^2}=\frac{x^2}{y^2}-2\frac{x}{y}\frac{t}{x}+\frac{t^2}{x^2}+2\frac{t}{y}=(\frac{x}{y}-\frac{t}{x})^2+2\frac{t}{y}\ge2\frac{t}{y}=2\frac{\sqrt{d}}{\sqrt{b}}.$ $\space\space\space\space\space\space\space\space (1)$

Now, if we replace $a, b, d$ in $(1)$ with respectively $c,d,b$, we'll get

$\frac{c}{d}+\frac{b}{c}\ge 2\frac{\sqrt{b}}{\sqrt{d}}$.

Next, use $(1)$ again, using numbers $\sqrt{b},\sqrt{d},\sqrt{d}$ and you'll get the desired result.

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