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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

max8128
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    What exactly do you mean that "you have created "? –  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 https://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 01:23

1 Answers1

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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 $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$

Michael Rozenberg
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  • 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