Prove by induction: inequality Use induction to prove the inequality:
$$\frac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdot 6\cdots 2n}\leq \frac{1}{\sqrt{n+1}}$$
I tried by multiplying $\frac{(2n+1)}{(2n+2)}$ to both sides but I don't know how to get $\frac{1}{\sqrt{n + 2}}$ on the RHS.
 A: Hint:  For positive integers $n$ one has
$\frac{1}{\sqrt{n+1}}\cdot\frac{(2n+1)}{(2n+2)}\leq \frac{1}{\sqrt{n+2}}$
is true if and only if
$\frac{1}{n+1}\cdot \frac{4n^2+4n+1}{4n^2+8n+4}\leq \frac{1}{n+2}~~~~~$   (seen by squaring both sides)
is true if and only if
$(n+2)(4n^2+4n+1)\leq (n+1)(4n^2+8n+4)~~~~~$   (seen by cross multiplying)
A: We know that the inequality holds for n=1. Now let us assume the inequality holds for n, and prove it for value n+1. The LHS and RHS of the inequality at n+1 is divided by the LHS and RHS of the inequality at n, respectively. We see that the LHS is multiplied by (2n+1)/(2n+2) and the RHS by [(n+1)^(1/2)]/[(n+2)^(1/2)]. Now, to prove the inequality, we have to prove:
(2n+1)/(2n+2)  ≤  √(n+1)/√(n+2)
((2n+1)^2)/((2n+2)^2)  ≤(n+1)/(n+2)
4n^3+12n^2+9n+2 ≤ 4n^3+12n^2+12n+4
Hence, proved.
A: We need to prove (if possible) that $\frac 1 {\sqrt {n+1}} *\frac{2n+1}{2n+2} \le \frac 1{\sqrt{n+2}}$
Which is true iff $\sqrt{n+2}(2n+1) \le \sqrt{n+1}(2n+2)$ which ... can be easily verified.
