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
10
$\begingroup$
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