# Inequality $\frac{\ln(7a+b)}{7a+b}+\frac{\ln(7b+c)}{7b+c}+\frac{\ln(7c+a)}{7c+a}\leq \frac{3\ln(8\sqrt{3})}{8\sqrt{3}}$

I'm interested by the following problem :

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

I have tried to use convexity and Jensen but the result is weaker . I try also Karamata's inequality but it fails totaly so I'm a bit lost .

If you have a hint it would be great .

Thank you .

• Such symmetric inequalities always can be solved by taking $a=b=c$. in this case we can derive $a=\sqrt{3}$. Aug 10, 2019 at 15:49
• source of problem please? Aug 10, 2019 at 17:16
• @C.F.G: A symmetric expression does not necessarily attains its maximum (or minimum) at a point where all variables are equal. Aug 11, 2019 at 4:41
• @MartinR, if maximum (or minimum) exist then the symmetric expression attains its maximum (or minimum) at a point where all variables are equal. isn't? please give a counterexample? what happens for the answer of this question? has been deleted? Aug 11, 2019 at 9:34
• @C.F.G: $f(x, y) = \exp((x+y)^2-(x+y)^4)$ is symmetric in $x, y$. The maximum is not attained on the line $x=y$. – The answer was deleted by its owner (after someone pointed out that it was wrong). Aug 11, 2019 at 10:04

Using the fact that $$\ln x \le \ln (8\sqrt{3}) + \frac{1}{8\sqrt{3}}(x-8\sqrt{3})$$ for $$x > 0$$ (the proof is simple and thus omitted), it suffices to prove that $$\sum_{\mathrm{cyc}}\frac{\ln (8\sqrt{3}) + \frac{1}{8\sqrt{3}}(7a+b-8\sqrt{3})}{7a+b} \le \frac{3\ln (8\sqrt{3})}{8\sqrt{3}}$$ or $$\frac{1}{7a+b} + \frac{1}{7b+c} + \frac{1}{7c+a} \le \frac{\sqrt{3}}{8}.$$ This has been solved in the link below: If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$
As pointed out by River Li, first show that for every $$x>0$$ holds the inequality $$\log x\leq\frac{x}{8\sqrt{3}}+\log\left(\frac{8\sqrt{3}}{e}\right).\tag 1$$ The equality in $$(1)$$ for holds only for $$x=8\sqrt{3}$$. From $$(1)$$ we have $$\sum_{cyc}\frac{\log(7a+b)}{7a+b}\leq \frac{\sqrt{3}}{8}+\sum_{cyc}\frac{1}{7a+b}\log\left(\frac{8\sqrt{3}}{e}\right).$$ Set $$\Pi(x,y,z):=\sum_{cyc}\frac{1}{7x+y},$$ Consider the inequality $$\frac{1}{kx+ly}\leq \frac{A}{kx}+\frac{B}{ly},$$
with $$k,l,x,y>0$$, $$k=7$$, $$l=1$$ and assume that $$\frac{A}{k}+\frac{B}{l}=\frac{1}{k+l}.$$ Hence $$B=\frac{1}{56}(7-8A)$$. But for this value of $$B$$ it is $$\frac{1}{kx+ly}-\frac{A}{7x}-\frac{B}{y}=-\frac{(x-y)(49x-56Ax-8Ay)}{56xy(7x+y)}.$$ Using $$49-56A=8A\Leftrightarrow A=\frac{49}{24}$$, we get $$\frac{1}{7x+y}-\frac{7}{64x}-\frac{1}{64y}=-\frac{7(x-y)^2}{64xy(7x+y)}\tag 2$$ Summing all three equalities $$(2)$$ (for the pairs $$(x,y),(y,z),(z,x)$$) we get $$\sum_{cyc}\frac{1}{7x+y}\leq\frac{7}{64}\sum_{cyc}\frac{1}{x}+\frac{1}{64}\sum_{cyc}\frac{1}{x}$$ and consequently $$\Pi(x,y,z)\leq\frac{1}{8}\sum_{cyc}\frac{1}{x}.$$ This is a ''bad'' estimate, since $$\sum_{cyc}\frac{1}{x}\geq\sqrt{3}.$$ However, I think, someone can work better with $$(2)$$ and obtain a ''good'' estimate.