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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 .

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  • $\begingroup$ Such symmetric inequalities always can be solved by taking $a=b=c$. in this case we can derive $a=\sqrt{3}$. $\endgroup$ – C.F.G Aug 10 at 15:49
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    $\begingroup$ source of problem please? $\endgroup$ – C.F.G Aug 10 at 17:16
  • $\begingroup$ @C.F.G: A symmetric expression does not necessarily attains its maximum (or minimum) at a point where all variables are equal. $\endgroup$ – Martin R Aug 11 at 4:41
  • $\begingroup$ @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? $\endgroup$ – C.F.G Aug 11 at 9:34
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    $\begingroup$ @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). $\endgroup$ – Martin R Aug 11 at 10:04
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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 $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$

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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.

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