Verifying an inequality I have a problem proving an inequality regarding probabilities.
You may prefer to skip to the definitions and the inequality right away 
without reading the paragraph below.
Suppose there are $n$ light bulbs. Independently, each light bulb is ON with probability $x$ and OFF with probability $(1-x)$. After the random variables realized, i.e., after it is observed which bulbs are on or off, one is smashed and this one is selected randomly among those that are OFF.
Let 
$$\alpha = x^{n-2}$$
be the probability that, given two lights $i$ and $j$ are OFF, all other lights are ON.
Let 
$$\beta_i = \sum\limits_{i=0}^{n-2} {n-2 \choose i}\frac{1}{i+1} x^{n-2-i}(1-x)^i = \frac{1-x^{n-1}}{(n-1)(1-x)}$$
be the probability that, given light $i$ is off and light $j$ is on, light bulb $i$ is smashed.
Similarly, define $\beta_j = \beta_i$.
Let 
$$\beta_{i,j} = \sum\limits_{i=0}^{n-2} {n-2 \choose i}\frac{2}{i+2} x^{n-2-i}(1-x)^i$$
be the probability that either $i$ or $j$ are smashed, given both of them are OFF.
It should hold that
$$ (1-\alpha) (1-\beta_{i,j}) \geq (1-\beta_i) (1- \beta_j) \quad  \forall n > 2, x\in [0,1]  $$
Unfortunately, I am not able to show that this generally holds. It would be great to point me to the basic math that I am missing.
 A: One might as well assume $i=1$ and $j=2$. Let $Y$ be the bulb randomly chosen to be smashed and let $\bar{S}_n$ denote the number of bulbs off among the bulbs $\{2,3,4,...,n\}.$ Let $A_0$ be the event $\{X_1=0,X_2=1\}$ and $A_1 = \{ X_1 = X_2 =0\}.$ You would like to show:
$$
P(Y \neq 1 | A_0)^2 \le P(\bar{S}_n > 0 | A_1) P(Y \notin \{1,2\} | A_1) ~~~(*).
$$
[Below $x$ denotes the probability that a bulb goes off] 
If we partition both sides on the values that $\bar{S}_n$ can take, the last inequality is the same as:
$$
\sum_{j,k} P(Y \neq 1, \bar{S_n}=j | A_0)
P(Y \neq 1, \bar{S_n}=k | A_0)
 \le 
\sum_{j,k}
P(\bar{S}_n =j | A_1) P(Y \notin \{1,2\}, \bar{S}_n = k | A_1).
$$
Fix $j \neq k$ and take the sum of the $(j,k)$ and $(k,j)$ terms on each side:
$$
2\frac{k}{k+1}\frac{j}{j+1}\binom{n-2}{j} \binom{n-2}{k} x^{j+k} (1-x)^{2n-4-(j+k)} 
$$
on the left and
$$
\left(\frac{k}{k+2} + \frac{j}{j+2}\right)\binom{n-2}{j}\binom{n-2}{k} x^{j+k} (1-x)^{2n-4-(j+j)}  
$$
on the right.
It turns out that 
$$
2\frac{k}{k+1}\frac{j}{j+1} \le \frac{k}{k+2} + \frac{j}{j+2}
$$
for all $j,k \in \{1,2,3,..,n-2\}^2$ $j \neq k.$ Thus we see that the sum of each pair of cross terms on the left is dominated by the corresponding pair on the right. One does a similar comparison for the square terms. Summing over $j$ and $k$ over these inequalities one obtains $(*).$
