# Inequality with power and condition

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 .

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