How to prove this inequality: $\sum_{cyc} \frac {1}{\alpha + \log_a {b}} \le \frac {2}{\alpha}$? This is another Olympiad problem my teacher gave us to train.
$$\sum_{cyc} \frac {1}{\alpha + \log_a{b}} \le \frac {2}{\alpha}$$ with $\alpha \in (0, 2]$ and $a, b, c \in (0, 1)$ or $ a, b, c \in(1, \infty)$. After working a little bit with it, we get $$\sum_{cyc} \log_{a^\alpha b} {a} \le \frac {2}{\alpha}$$ but I don't know how to go on.

Starting from @Robert Z's answer, consider that we look for the maximum value of function
$$f(x,y,z)=\frac {1}{\alpha + x}+\frac {1}{\alpha + y}+\frac {1}{\alpha + z}$$ Using the constraint $xyz=1$, we look at the maximum of function
$$g(x,y)=\frac{2 \alpha +x^2 y (2 \alpha +y)+x \left(2 \alpha  y^2+3 \alpha ^2
   y+1\right)+y}{(\alpha +x) (\alpha +y) (\alpha  x y+1)}$$ Computing the partial derivatives
$$\frac {\partial g(x,y)}{\partial x}=\frac{y}{(\alpha  x y+1)^2}-\frac{1}{(\alpha +x)^2}=0$$
$$\frac {\partial g(x,y)}{\partial y}=\frac{x}{(\alpha  x y+1)^2}-\frac{1}{(\alpha +y)^2}=0$$
The only real solutions are
$$x=1 \quad y=1\quad z=1 \implies f(x,y,z)=\frac{3}{\alpha +1}$$
$$x=\frac 1{\alpha^2}\quad y=\alpha^4\quad z=\frac 1{\alpha^2}\implies f(x,y,z)=\frac{2 \alpha ^3+1}{\alpha(\alpha ^3+1) }$$
$$x={\alpha^2}\quad y=\alpha^2\quad z=\frac 1{\alpha^4}\implies f(x,y,z)=\frac{2 \alpha ^3+1}{\alpha(\alpha ^3+1) }$$
$$x={\alpha^4}\quad y=\frac 1{\alpha^2}\quad z=\frac 1{\alpha^2}\implies f(x,y,z)=\frac{2 \alpha ^3+1}{\alpha(\alpha ^3+1) }$$
So, if $\alpha=2$, the maximum value is $\frac 12$
If $0 <\alpha<2$, the maximum value is $\frac 2 \alpha$
If $\alpha>2$, the maximum value is $\frac 3 {\alpha+1}$
