<s>[INST] 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 $p-q$.
 [/INST] The first one can also be slightly generalized. If $c>=0$ then:
$$
\frac{p+c}{q+c} + \frac{q+c}{p+c} \ge 1 + \frac{p}{2q+c} + \frac{q}{2p+c} + \frac{c}{p+q+c}
$$
The proof is the same as suggested in my comment above.
</s>