How would one show that for positive $a,b,c,d$ and $a+b+c+d = 4$ that $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \leq \frac{4}{abcd} $$
Another symmetric inequality
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2Did you try some examples (e.g. $a=b=c=d=10$)? – 2012-02-02
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0$$\sqrt{ab} \leq \frac{a+b}{2}$$ – 2012-02-02
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0Perhaps 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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0Note that $a+b+c+d = 4$ – 2012-02-02
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0@Robert: $10+10+10+10\ne4$. Also, both sides are homogeneous, just not with the same degree. – 2012-02-02
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1@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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0@JamesGayson,Rewrite LHS into form of one fraction... – 2012-02-02
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0Yes, 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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0I don't know how I missed the $a+b+c+d=4$. – 2012-02-02
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0@JamesGayson,I have proved that numerator is certainly less than $16$ ... – 2012-02-02
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0Lagrange multipliers work well here :) – 2012-02-02
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3C'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
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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2I love this proof. Aryabhata I am your fan! – 2012-02-24
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0@KirthiRaman: You are very kind! Thanks! – 2012-02-24
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0I cannot resist to agree with the comments here. This is a really beautiful proof. – 2012-02-25
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 $a+b+c+d = 4$, $abcc+bcdd+cdaa+dabb \le 4$.
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:
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1$y$ is not a symmetric polynomial. – 2012-02-23
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0@mercio You are right. I just reminded me on them. I took it out. Thanks – 2012-02-24
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0the 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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0you are right. I took it back +1 for yours – 2012-02-24