# Inequality : $\exp\Big(\frac{1}{7a+b}\Big)+\exp\Big(\frac{1}{7b+c}\Big)+\exp\Big(\frac{1}{7c+a}\Big)\leq 3\exp\Big(\frac{1}{8\sqrt{3}}\Big)$

I'm interested by the following problem :

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

My try : I transform this inequality in a problem of tangent line on the exponential (because $$e^x=(e^x)'$$) and with the condition we have (it's checked) : $$f(a,b,c)\leq\sum_{cyc} \frac{\exp\Big(\frac{1}{7a+b}\Big)-\exp\Big(\frac{1}{8\sqrt{3}}\Big)}{\frac{1}{7a+b}-\frac{1}{8\sqrt{3}}}\leq 3\exp\Big(\frac{1}{8\sqrt{3}}\Big)$$

After this I want to use the three chord lemma but it gives not enought informations .

Maybe we can use the Lagrange multiplier but I don't control this method .

If you have hints it would be nice and thanks a lot for your time .

Ps: This is the exponential twin of this If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$

• This is wrong for $(a, b, c) = (1, 2, 3)$. – Martin R Jul 5 at 15:40
• @MartinR he has $a\ge b\ge c$ – Empy2 Jul 21 at 12:55
• @Empy2: That restriction has been added today. What I am missing is: Why should the inequality hold (in particular if it is stronger than the linked-to unsolved problem) and why is $a \ge b \ge c$ a sensible restriction (apart from invalidating the counter-example). – Martin R Jul 21 at 14:32