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.
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.
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 $n\ge1$.
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.