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I have finally found the right function such that we have :

Let $a,b,c>0$ such that $abc=a+b+c$ then we have : $$\sum_{cyc}\tan\Big(\frac{1}{7a+b}\Big)\leq 3\tan\Big(\frac{1}{8\sqrt{3}}\Big)$$

To underline the difficulty we have also :

Let $a,b,c>0$ such that $\pi=a+b+c$ and $a,b,c<\frac{\pi}{2}$then we have : $$\sum_{cyc}\tan\Big(\frac{1}{7\tan(a)+\tan(b)}\Big)\leq 3\tan\Big(\frac{1}{8\sqrt{3}}\Big)$$

It's more impressive like this and I don't know what to do with this in fact it's more a present for Michael Rozenberg .

I accept ugly proof but I think it's too hard to find .

Any hint would be appreciable .

Thanks in advance .

  • What does “we have” mean? You know that it is true, you assume that it is true, or you guess that it is true? – Martin R Jul 10 '19 at 15:15
  • @MartinR I have tested this inequality numerically and I assume it's true –  Jul 10 '19 at 15:17
  • The LHS of the second inequality is not defined if any of $a, b, c$ is $\pi/2$. – Martin R Jul 10 '19 at 15:21
  • @MartinR thanks I have corrected . –  Jul 10 '19 at 15:25
  • Note that your first inequality is stronger than https://math.stackexchange.com/q/2197138/42969, which has no (accepted) proof so far. I (personally) doubt that it will be easier to prove your version. – Martin R Jul 10 '19 at 15:27

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