<s>[INST] $\frac{(2n − 1)!!}{(2n)!!} \leq \frac{1}{\sqrt{2n+1}}$ 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. 
 [/INST] $$\frac{(2n-1)!!}{(2n)!!}=\frac{2n-1}{2n}\cdot \frac{(2n-3)!!}{(2n-2)!!}\le \frac{2n-1}{2n\sqrt{2(n-1)+1}}=\frac{2n-1}{2n\sqrt{2n-1}}= \frac{\sqrt{2n-1}}{2n}$$
Now to show that $\frac{\sqrt{2n-1}}{2n}\le \frac{1}{\sqrt{2n+1}}$ we show that $\frac{1}{\sqrt{2n+1}}-\frac{\sqrt{2n-1}}{2n}\ge 0$:
$$\frac{1}{\sqrt{2n+1}}-\frac{\sqrt{2n-1}}{2n}=\frac{2n-\sqrt{2n-1}\sqrt{2n+1}}{2n\sqrt{2n+1}}=\frac{2n-\sqrt{4n^2-1}}{2n\sqrt{2n+1}}\ge \frac{2n-\sqrt{4n^2}}{2n\sqrt{2n+1}}=0$$
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