Following is the inequation I have been trying to prove for a while.
$$\frac{\frac{1}{2}(1-q)}{\frac{1}{2}(1-q) + pq}\neq \frac{\frac{1}{4}(1-q)}{\frac{1}{4}(1-q) + p^2q} + \frac{\frac{1}{4}(1-q)}{\frac{1}{4}(1-q) + qp(1-p)} $$
Of course, I tried the basic cancellation out but the expressions only got bigger. It also occurred to me that I didn't how prove that their NOT equal. Any help would be appreciated.
Clearly the two expressions are equal when $q=1$ and $0<p<1$, so assume that $q\ne 1$; the inequality can then be simplified to
$$\frac1{1-q+2pq}\ne\frac1{1-q+4p^2q}+\frac1{1-q+4qp(1-p)}$$
and then, putting the righthand side over a common denominator, to
$$\frac1{1-q+2pq}\ne\frac{2-2q+4pq}{(1-q)^2+4pq(1-q)+16p^3q^2(1-p)}\;.$$
Assuming that the denominators are not zero, this holds if and only if
$$\begin{align*} (1-q)^2+4pq(1-q)+16p^3q^2(1-p)&\ne 2(1-q+2pq)^2\\ &=4(1-q)^2+16pq(1-q)+16p^2q^2\;, \end{align*}$$
or
$$3(1-q)^2+12pq(1-q)+16p^2q^2\big(1-p(1-p)\big)\ne 0\;.$$
This is clearly the case when $0\le p\le 1$ and $0<q<1$.