Hard inequality with condition ($xyz=1$)

It's a new inequality that I have created this is the following :

Let $$x,y,z$$ be real positive numbers such that $$xyz=1$$ then we have : $$\sum_{cyc}^{}\frac{(\frac{z}{x})^{\frac{1}{2}}}{(z+1+\frac{1}{x})^{\frac{1}{2}}(7+\frac{1}{x})}\leq \frac{\sqrt{3}}{8}$$

I have tried to apply Jensen's inequality to the function $$f(x)=\frac{1}{7+\frac{1}{x}}$$ wich is concave but we don't get a good expression . I also tried to expand the expression but it fails . After this I have no idea to prove this...

Any hints would be appreciable. Thanks in advance

• What exactly do you mean that "you have created "? – complexmanifold Oct 27 '18 at 15:03
• Is this related to an actual mathematical contest, as described in the contest-math tag wiki? In that case it would be appropriate to add a source to the contest (compare math.meta.stackexchange.com/a/28999/42969). Otherwise you can remove the tag. – Martin R Oct 27 '18 at 15:26
• The variable $y$ does not explicitly appear in your left-hand side. Perhaps you intend for the "cyclic" summation to be taken in a particular order, but with only $x,z$ shown, it may be slightly ambiguous which three(?) terms are meant to be added. – hardmath Oct 30 '18 at 1:23

Let $$x=\frac{a}{b}$$ and $$y=\frac{b}{c},$$ where $$a$$, $$b$$ and $$c$$ are positives.

Thus, $$z=\frac{c}{a}$$ and we need to prove that $$\sum_{cyc}\frac{1}{7a+b}\leq\frac{1}{8}\sqrt{\frac{3(a+b+c)}{abc}},$$ which was here:

If $$a+b+c=abc$$ then $$\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$$

• But that other inequality has no proof either. – My guess would be that this question was posted as an attempt to solve your older problem. – Martin R Oct 27 '18 at 16:29