Another symmetric inequality

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. $a=b=c=d=10$)? –  Robert Israel Feb 2 '12 at 19:30
$$\sqrt{ab} \leq \frac{a+b}{2}$$ –  pedja Feb 2 '12 at 19:34
Perhaps there's some extra condition. Otherwise, since the left side is homogeneous and the right is not, this makes no sense. –  Robert Israel Feb 2 '12 at 19:36
@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. –  James Gayson Feb 2 '12 at 19:40
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. –  James Gayson Feb 2 '12 at 21:22

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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+1 very nice and stylish –  draks ... Feb 24 '12 at 9:53
I love this proof. Aryabhata I am your fan! –  Kirthi Raman Feb 24 '12 at 13:57
@KirthiRaman: You are very kind! Thanks! –  Aryabhata Feb 24 '12 at 15:11
I cannot resist to agree with the comments here. This is a really beautiful proof. –  Dejan Govc Feb 25 '12 at 0:01

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 $0<a<2$ and $c=d=1$. Substituting this, gives $$\begin{eqnarray*} (a-2)(a-1)^2&<&0 \end{eqnarray*}$$ which is true for the given range of $a$.

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$y$ is not a symmetric polynomial. –  mercio Feb 23 '12 at 21:43
@mercio You are right. I just reminded me on them. I took it out. Thanks –  draks ... Feb 24 '12 at 7:14
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. –  mercio Feb 24 '12 at 13:05
you are right. I took it back +1 for yours –  draks ... Feb 24 '12 at 13:17