Of course $x=\log_a(b)=\frac{\log b}{\log a}$ and $y=\log_b(a)=\frac{\log a}{\log b}$ so we have $xy=1$. Moreover, we only know that $x,y>0$. So we simply need to find the infimum of $\frac{1}{3+x}+\frac{1}{3+y}$ where $x,y>0$ and $xy=1$. Letting $x \to \infty, y \to 0$, the expression comes arbitrarily close to $\frac{1}{3}$ so $c \le \frac{1}{3}$. Let us now prove that $c=\frac{1}{3}$ i.e. that we have \[\frac{1}{3+x}+\frac{1}{3+y}>\frac{1}{3}\]for all $x,y>0$ with $xy=1$. In fact by expanding that inequality is immediately equivalent to $xy<9$ which is certainly satisfied since $1<9$. Done.

gobathegreat wrote: Let $a$ and $b$ be real numbers bigger than $1$. Find maximal value of $c \in \mathbb{R}$ such that $$\frac{1}{3+\log _{a} b}+\frac{1}{3+\log _{b} a} \geq c$$ Solution of Xiongchangjin:

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@above: What you proved is that the maximum of the expression is $\frac{1}{2}$. However, the problem clearly asks for the minimum value of the expression. So you are correct in claiming that $c \le \frac{1}{2}$ but that you could have done much easier by choosing $a=b=42$. This however is only remotely related to the problem since it does not prove that the inequality is true for $c=\frac{1}{2}$. And in fact it isn't as the example $a=b^2$ (or in fact any other choice with $a \ne b$) shows.