Three inequalities with sums of fractions over two positive integers In a proof, I arrive at three inequalities for all $p,q \geqslant 0$:
\begin{align}
\frac{p+1}{q+1} + \frac{q+1}{p+1} &\geqslant 1 +
      \frac{p}{2q+1} + \frac{q}{2p+1} + \frac{1}{p+q+1};\cr
\frac{2q+3}{p+1} + \frac{p+2}{q+2} &\geqslant 2 +
      \frac{2q+1}{2p+1} + \frac{2}{p+q+2};\cr
\frac{q+1}{p+1} + \frac{q+2}{p+2} + \frac{p+2}{q+2} +
      \frac{p+1}{q+1} &\geqslant 2 + \frac{2p+1}{2q+2} +
      \frac{2q+1}{2p+2} + \frac{2}{p+q+2}.
\end{align}
Any idea on how to attack these?
EDIT: Following the pieces of advice in comments, I expanded everything to get rid of the fractions and form bivariate polynomials which must be positive. In the first case, there is an obvious factor $pq$. Since the polynomial is zero if $p=q$ (the inequality is tight), this means that $p-q$ is another factor, yielding $pq(p-q)^2(2p+2q+3) \geqslant 0$. In the case of the second polynomial, there is a trivial factor $q$, and, again $p-q$. I didn't know how to guess the two last factors, but Wolfram|Alpha helped: $q(p-q)(p-q-1)(2p+2q+5)
    \geqslant 0$.
The last one is fearsome, although the polynomial must have a factor $p-q$.

For the last one, multiply through by $2(p+1)(q+1)(p+2)(q+2)(p+q+2)$ to get rid of the fractions; then Wolfram|Alpha factors the difference between the left-hand side and the right-hand side as
$$
(p-q)^2\left(2p^2(q+1)+p(2q^2+9q+8)+2(q+2)^2\right)\;,
$$
which is $\ge0$ for $p,q\ge0$.
