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.
 A: First we note that in the statement, it should be
$$a, b, c \in (0, 1) \text{ OR } a, b, c \in (1, \infty) \qquad (\star).$$
Indeed for $\alpha=2$, $a=b=1/2$ and $c=2$ the inequality does not hold:
$$\frac {1}{2 + 1}+\frac {1}{2  -1}+\frac {1}{2 -1}=\frac{7}{3}> 1=\frac {2}{2}.$$
Assuming $(\star)$, let $x=\log_a{b}$, $y=\log_b{a}$, $z=\log_c{a}$, then $x,y,z>0$, $xyz=1$ and we have to show that
$$\frac {1}{\alpha + x}+\frac {1}{\alpha + y}+\frac {1}{\alpha + z} \le \frac {2}{\alpha}.$$
The above inequality is equivalent to
$$\frac{\alpha^3-2}{\alpha}\leq xy+yz+zx.$$
Since $\max_{\alpha\in(0,2]}\frac{\alpha^3-2}{\alpha}=\frac{(2)^3-2}{2}=3$, it suffices to show that
$$3\leq xy+yz+zx$$
which directly follows by AM-GM.
A: As Robert Z said, the condition should be $\{a,b,c\}\subset(0,1)$ or $\{a,b,c\}\subset(1,+\infty)$,
otherwise your inequality is wrong.
Let $\ln{a}=x$, $\ln{b}=y$ and $\ln{c}=z$.
Thus, we need to prove that:
$$\sum_{cyc}\frac{1}{\alpha+\frac{\ln{b}}{\ln{a}}}\leq\frac{2}{\alpha}$$ or
$$\sum_{cyc}\frac{1}{\alpha+\frac{y}{x}}\leq\frac{2}{\alpha}.$$
Now, by the conditions any fractions as $\frac{\ln{a}}{\ln{b}}$ are positives and it's enough to assume that $x$, $y$ and $z$ are positives and we need to prove that
$$\sum_{cyc}\frac{x}{\alpha x+y}\leq\frac{2}{\alpha}$$ or
$$\sum_{cyc}\left(\frac{x}{\alpha x+y}-\frac{1}{\alpha}\right)\leq-\frac{1}{\alpha}$$ or
$$\sum_{cyc}\frac{y}{\alpha x+y}\geq1,$$ which is true by C-S:
$$\sum_{cyc}\frac{y}{\alpha x+y}=\sum_{cyc}\frac{y^2}{\alpha xy+y^2}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(\alpha xy+y^2)}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(2xy+y^2)}=1.$$
A: 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 $\frac 3 {\alpha+1}$
