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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 $p-q$.

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    $\begingroup$ Just a hint. Multiply both sides of the first by $$(p + 1) (q + 1) (2 q + 1) (2 p + 1) (p + q + 1)$$ to eliminate all fractions. Then factor to get $$p (p - q)^2 q (3 + 2 p + 2 q)>=0$$ $\endgroup$ – ivan Sep 11 '12 at 12:15
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    $\begingroup$ In general, you can reach the simplest form by adding and multiplying everything together using brute force. It may look hard, but really, it just takes some time. $\endgroup$ – Karolis Juodelė Sep 11 '12 at 12:19
  • $\begingroup$ Thanks, Ivan and Karolis. It took me almost an hour to get the first one right by hand, but nothing really hard. $\endgroup$ – Christian Sep 11 '12 at 13:38
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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 $p,q\ge0$.

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  • $\begingroup$ Thank you, joriki. I have a theorem now! $\endgroup$ – Christian Sep 12 '12 at 5:51
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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.

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