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Given a positive integer $n$, is there a simple way to see that $$ (n+3)^{n+2}(2n+5)^{n+2}(n+1)^{n+2}(2n+1)^n \ge (n+2)^{2n+4}(2n+3)^{2n+2} $$

Any hint is welcome.

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Where does it come from? –  Victor Mar 17 '12 at 1:19
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-1: Your question seems to show no thought of your own. –  JavaMan Mar 17 '12 at 1:22
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@Victor: Are you looking for a repository of questions that haven't yet been asked on this site? –  The Chaz 2.0 Mar 17 '12 at 1:35
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I think I can replace the $\ge$ by $\gt$. –  André Nicolas Mar 17 '12 at 1:36

2 Answers 2

up vote 2 down vote accepted

Multiply both sides by $(2n+1)^2$, and raise both sides to the power $1/(n+2)$.

On the left, you get $$(n+3)(2n+5)(n+1)(2n+1)$$ which is $$4n^4+28n^3+65n^2+56n+15$$ On the right, you get $$(n+2)^2(2n+3)^2\left({2n+1\over2n+3}\right)^{2/(n+2)}$$ which is $$(4n^4+28n^3+73n^2+84n+36)\left(1+{2\over2n+1}\right)^{-2/(n+2)}$$ So we want to show $$(4n^4+28n^3+65n^2+56n+15)\left(1+{2\over2n+1}\right)^{2/(n+2)}\ge4n^4+28n^3+73n^2+84n+36$$ Now $$\left(1+{2\over2n+1}\right)^{2/(n+2)}\ge1+{4\over(2n+1)(n+2)}={2n^2+5n+6\over2n^2+5n+2}$$ so we win if $$(4n^4+28n^3+65n^2+56n+15)(2n^2+5n+6)\ge(4n^4+28n^3+73n^2+84n+36)(2n^2+5n+2)$$ If my arithmetic is correct, the first three terms match, while the coefficient of $n^3$ is 605 on the left side, 589 on the right. Maybe someone wants to check my work and complete the calculation. If, as I suspect, each coefficient on the left is at least as big as the corresponding coefficient on the right, we're done. If some coefficient on the right exceeds the corresponding coefficient on the left, we still have some work to do to prove that some cubic is never negative for $n\ge1$.

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Try using the fact that $(n+3)^{n+2}\geq (n+2)^{n+2}$ and $(2n+5)^{n+1}>(2n+3)^{n+1}$. Then you only need to show that $$(2n+5)(n+1)^{n+2}(2n+1)^n \ge (n+2)$$ which is obvious.

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Sorry, I mistype the question, it should be $$ (n+3)^{n+2}(2n+5)^{n+2}(n+1)^{n+2}(2n+1)^n \ge (n+2)^{2n+4}(2n+3)^{2n+2} $$ –  Sunni Mar 17 '12 at 13:37

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