I have this to propose :

Let $a,b,c>0$ such that $abc=a+b+c$ and $a\geq b \geq c$ then we have : $$\sum_{cyc}\frac{a}{5a+c+1}\leq \frac{3(3)^{0.5}}{6(3)^{0.5}+1}$$

I can prove the following inequality under the same assumptions : $$\sum_{cyc}\frac{a}{5a+c} \leq 0.5$$ Wich is not so hard. But the add of the coefficient 1 disturbed me. So if you have nice ideas it world cool... Thanks!

  • $\begingroup$ Appearance of $3\sqrt 3$ with those coefficients makes me feel some Cauchy-Schwarz kind of way $\endgroup$ – Certainly not a dog Oct 2 at 13:58

Since $$\sqrt{\frac{abc}{a+b+c}}\geq\frac{\sqrt3abc}{ab+ac+bc}$$ it's $$(ab+ac+bc)^2\geq3abc(a+b+c)$$ or $$\sum_{cyc}c^2(a-b)^2\geq0,$$ we obtain: $$\sum_{cyc}\frac{a}{5a+c+1}=\sum_{cyc}\frac{a}{5a+c+\sqrt{\frac{abc}{a+b+c}}}\leq\sum_{cyc}\frac{a}{5a+c+\frac{\sqrt3abc}{ab+ac+bc}}\leq\frac{3\sqrt3}{6\sqrt3+1},$$ where the last inequality is obviously true by BW: https://artofproblemsolving.com/community/c6h522084


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