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How would one show that for positive $a,b,c,d$ and $a+b+c+d = 4$ that $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \leq \frac{4}{abcd} $$

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    Did you try some examples (e.g. $a=b=c=d=10$)?2012-02-02
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    $$\sqrt{ab} \leq \frac{a+b}{2}$$2012-02-02
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    Perhaps there's some extra condition. Otherwise, since the left side is homogeneous and the right is not, this makes no sense.2012-02-02
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    Note that $a+b+c+d = 4$2012-02-02
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    @Robert: $10+10+10+10\ne4$. Also, both sides are homogeneous, just not with the same degree.2012-02-02
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    @pedja, yes I've been playing around this that sort of thing but couldn't make it work. Can you be more explicit? Clearly also the average of the four numbers is 1 and hence $abcd \leq 1$. But even so, I'm still stuck.2012-02-02
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    @JamesGayson,Rewrite LHS into form of one fraction...2012-02-02
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    Yes, even so $$ LHS = \frac{a^2cd + b^2da + c^2ab + d^2bc}{abcd}$$ How do we show the numerator is $\leq 4$?2012-02-02
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    I don't know how I missed the $a+b+c+d=4$.2012-02-02
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    @JamesGayson,I have proved that numerator is certainly less than $16$ ...2012-02-02
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    Lagrange multipliers work well here :)2012-02-02
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    C'mon people. I'm not a 15 year old in the middle of an exam. Give me a constructive hint or better yet, show a complete solution. I also have the Lagrange multiplier solution, but I think it's too inelegant. I'm looking for something more stylish.2012-02-02

3 Answers 3

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Consider

$$\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right)abcd = a^2cd + b^2ad + c^2ab + d^2bc = ac(ad + bc) + bd(ab + cd)$$

Since there is cyclic symmetry, we can assume that $ad + bc \le ab + cd$.

So

$$ac(ad + bc) + bd(ab + cd) \le (ac + bd)(ab + cd)$$

Now $xy \le \left(\frac{x+y}{2}\right)^2$

and so

$$(ac + bd)(ab + cd) \le \left(\frac{ac + bd + ab + cd}{2}\right)^2 = \left(\frac{(a+d)(b+c)}{2}\right)^2$$

Applying $xy \le \left(\frac{x+y}{2}\right)^2$ again we get

$$\left(\frac{(a+d)(b+c)}{2}\right)^2 \le \left(\frac{\left(\frac{a+b+c+d}{2}\right)^2}{2}\right)^2 = 4$$

Thus $$\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a}\right)abcd \le 4$$

and so

$$\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \le \frac{4}{abcd}$$


What we have shown is that, for four positive numbers,

$$ \left(\frac{a+b+c+d}{4}\right)^4 \ge abcd\frac{\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{d}+ \frac{d}{a}\right)}{4}$$

and since $\frac{a}{b} + \frac{b}{c} + \frac{c}{d}+ \frac{d}{a} \ge 4$, this inequality is stronger than $\text{AM} \ge \text{GM}$ for $4$ numbers.

Somewhat surprisingly, we only used $\text{AM} \ge \text{GM}$ (twice) to prove it! And for two numbers, a similar inequality is actually false!

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    I love this proof. Aryabhata I am your fan!2012-02-24
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    @KirthiRaman: You are very kind! Thanks!2012-02-24
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    I cannot resist to agree with the comments here. This is a really beautiful proof.2012-02-25
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Aryabhata's nice proof can be restated as :

$$ (ac+bd)((a+b+c+d)^4 - 64(abcc+bcdd+cdaa+dabb)) \\ =ac(16(ac+bd-ad-bc)^2+(a+b-c-d)^2((a+b+c+d)^2+4(a+b)(c+d))) \\ +bd(16(ac+bd-ab-cd)^2+(b+c-d-a)^2((a+b+c+d)^2+4(b+c)(d+a))) \\ \ge 0$$

Therefore, if $a+b+c+d = 4$, $abcc+bcdd+cdaa+dabb \le 4$.

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Let's assume $a,b,c,d>0$. Rewriting your equation gives: $$ \begin{eqnarray*} a^2cd+b^2ad+c^2ab+d^2bc\leq 4 \end{eqnarray*} $$ Equality is reached, if $a=b=c=d=1$. It's left to show, that this maximal:

Let $b=(2-a)$ with $0

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    $y$ is not a symmetric polynomial.2012-02-23
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    @mercio You are right. I just reminded me on them. I took it out. Thanks2012-02-24
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    the point was that it is not possible to express $y$ as a polynomial in $e_1,e_2,e_3,e_4$ like you did.2012-02-24
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    you are right. I took it back +1 for yours2012-02-24