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I have this to propose :

Let $a,b,c,d$ be real positive numbers such that $abcd=1$ then we have : $$\sum_{cyc}a^{ab}\geq 4$$

First I definitively can't prove this by my own but if someone can prove this it would be very helpful to demonstrate this : Prove that $a^{ab}+b^{bc}+c^{cd}+d^{da} \geq \pi$ . If my result is right furthermore we can have the precision wanted to approach the minimum . It will be enough to work on the two conditions $a+b+c+d=4$ and $abcd=\alpha$ to have the conclusion .

Edit : First thanks to MartinR to underline my mistakes , secondly in fact the inequality works for some $\alpha$ with $abcd=\alpha$ and $0<\alpha$ but I don't know further . So I prefer restrict the $\alpha$ to one .

Thanks in advance .

  • I tried some random values, and (unless I made some error) the inequality does not hold e.g. for $(a, b, c, d) = (0.1, 0.1, 0.3, 1.0)$. – Martin R Dec 21 '18 at 21:55

1 Answers1

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Here I have proved this

$$n+\sum_{cyc}\ln(a_i^{a_i a_{i+1}})\leq \sum_{cyc}a_i^{a_i a_{i+1}}$$

It's easy to conlude if we note that with $\prod_{i=1}^{n}a_i=1$ we have : $$\sum_{cyc}\ln(a_i^{a_i a_{i+1}})\geq 0$$

Now put $n=4$ and we have your result .

max8128
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