<s>[INST] Help with this proof by induction with inequalities. Show that mathematical induction can be used to
prove the stronger inequality $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} < \frac{1}{\sqrt{3n + 1}}$ for all integers greater than 1, which, together with a verification for the case where n = 1, establishes the weaker inequality we originally tried to prove using mathematical induction.
Base Case: P(2)
\begin{aligned}
\frac{1}{2}\cdot\frac{2(2)-1}{2(2)} &< \frac{1}{\sqrt{3(2) + 1}}\\
\frac{3}{8} &< \frac{1}{\sqrt{7}}\\
\frac{1}{8} &< \frac{1}{3\sqrt{7}}\\
\end{aligned}
This is true as $8 > 3\sqrt{7}$.
Inductive Hypothesis: $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} < \frac{1}{\sqrt{3n+1}}$
In the inductive step, we want to show that $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} \cdot \frac{2n+1}{2n+2} < \frac{1}{\sqrt{3n+4}}$.
Using the inductive hypothesis, we can get to the following:
\begin{aligned}
\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} \cdot \frac{2n+1}{2n+2} &< \frac{1}{\sqrt{3n+1}}\cdot \frac{2n+1}{2n+2}\\
&< \frac{1}{\sqrt{3n+1}}\cdot 1\\
\end{aligned}
I am not sure how to get to $< \frac{1}{\sqrt{3n+4}}$ from here because i know that    if the denominator would get bigger by adding 3 to it so the inequality wouldn't follow...
 [/INST] You need to show that $$\frac{1}{\sqrt{3n+1}}\cdot \frac{2n+1}{2n+2}\\
< \frac{1}{\sqrt{3n+4}}\\$$Square both sides and cross multiply to get $$ (2n+1)^2 (3n+4)<(3n+1)(2n+2)^2$$
Multiply and cancel equal terms to get $$ 12n^2+12n+3 <12n^2 +13n+3 $$
Which is true for all positive integers.
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