$\frac{(2n − 1)!!}{(2n)!!} \leq \frac{1}{\sqrt{2n+1}}$ by induction? Let $(2n)!!$ be the product of all positive even integers less than or equal to $2n$. Let $(2n − 1)!!$ be the product of all odd positive integers less than or equal to $(2n − 1)$. Prove that 
$$\frac{(2n − 1)!!}{(2n)!!} \leq \frac{1}{\sqrt{2n+1}}.$$
I am up to the inductive step, where am I stuck. I can't find a way to substitute in from the assumption because of the square root. 
 A: This is not an answer, but I'll help you to write this in a convenient form:

Note that$$(2n)!!=2\times4\times\cdots\times(2n-2)\times(2n)=2^nn!$$  $$(2n-1)!!=1\times3\times\cdots\times(2n-3)\times(2n-1)=\dfrac{(2n-1)!}{2^{n-1}(n-1)!}.$$ 

Therefore $$\dfrac{(2n-1)!!}{(2n)!!}=\dfrac{\dbinom{2n-1}{n}}{2^{2n-1}}.$$ In fact using mathematical induction you can prove that for $n\gt 0,$$$\dfrac{1}{2n+1}\lt\dfrac{\dbinom{2n-1}{n}}{2^{2n-1}}\lt\dfrac{1}{\sqrt{2n+1}}.$$
A: You have to deduce from the inductive hypothesis, namely
$$\frac{(2n-1)!!}{(2n)!!}\le\frac1{\sqrt{2n+1}}$$
for some $n>0$ that
$$\frac{(2n+1)!!}{(2n+2)!!}\le\frac1{\sqrt{2n+3}}.$$
At any case you deduce that
$$\frac{(2n+1)!!}{(2n+2)!!}=\frac{(2n-1)!!}{(2n)!!}\,\frac{2n+1}{2n+2}\le \frac1{\sqrt{2n+1}}\,\frac{2n+1}{2n+2},$$
so it suffices to prove that
$$\frac1{\sqrt{2n+1}}\,\frac{2n+1}{2n+2}\le \frac{1}{\sqrt{2n+3}}\iff\sqrt{2n+1}\sqrt{2n+3}\le 2n+2.$$
It shouldn't be too hard to end the inductive step.
A: $$\frac{(2n-1)!!}{(2n)!!}=\frac{2n-1}{2n}\cdot \frac{(2n-3)!!}{(2n-2)!!}\le \frac{2n-1}{2n\sqrt{2(n-1)+1}}=\frac{2n-1}{2n\sqrt{2n-1}}= \frac{\sqrt{2n-1}}{2n}$$
Now to show that $\frac{\sqrt{2n-1}}{2n}\le \frac{1}{\sqrt{2n+1}}$ we show that $\frac{1}{\sqrt{2n+1}}-\frac{\sqrt{2n-1}}{2n}\ge 0$:
$$\frac{1}{\sqrt{2n+1}}-\frac{\sqrt{2n-1}}{2n}=\frac{2n-\sqrt{2n-1}\sqrt{2n+1}}{2n\sqrt{2n+1}}=\frac{2n-\sqrt{4n^2-1}}{2n\sqrt{2n+1}}\ge \frac{2n-\sqrt{4n^2}}{2n\sqrt{2n+1}}=0$$
