Show $\frac{3\cdot 7\cdot 11\cdot _{...}\cdot\left(4n-1\right)}{5\cdot9\cdot13\cdot _{...}\cdot \left(4n+1\right)\:}<\sqrt{\frac{3}{4n+3}}$ for $n>0$

Question

I need to prove that the expression: $$\frac{3\cdot 7\cdot 11\cdot _{...}\cdot \left(4n-1\right)}{5\cdot 9\cdot 13\cdot _{...}\cdot \left(4n+1\right)\:}<\sqrt{\frac{3}{4n+3}}$$ holds true for all n>0. What would be the simplest way to go about this?

So far I tried this:

Show it holds true for p(1): $$\frac{3}{5}<\sqrt{\frac{3}{7}}$$

which it does.

Next I assume it holds true for p(n) and the only step left is to prove for p(n+1): $$\frac{3\cdot 7\cdot 11\cdot _{...}\cdot \left(4n-1\right)\left(4n+3\right)}{5\cdot 9\cdot 13\cdot _{...}\cdot \left(4n+1\right)\:\left(4n+5\right)}<\sqrt{\frac{3}{4n+7}}$$

$$\frac{4n+3}{4n+5}\sqrt{\frac{3}{4n+3}}<\sqrt{\frac{3}{4n+7}}$$

$$\frac{4n+3}{4n+5}\sqrt{\frac{3}{4n+3}}-\sqrt{\frac{3}{4n+7}}<0$$

Here the expression get's kind of messy and it's "hard" to evaluate it, so is there a better and faster way?

Answer 1

From the second last step, it can be said that we have to prove that $$\frac{4n+3}{4n+5} \frac{1}{\sqrt{4n+3}} < \frac{1}{\sqrt{4n+7}}$$ or,$$\sqrt{(4n+3)(4n+7)} < (4n + 5) $$ or, since n>0, we can square both sides without change of signs$$16n^2 + 40n + 21 < 16n^2 + 40n + 25 $$ which is true for all $n > 0$. Hence, proved.

Answer 2

A few transformations starting at the line before the last line \begin{align*} \frac{4n+3}{4n+5}\sqrt{\frac{3}{4n+3}}<&\sqrt{\frac{3}{4n+7}}\\ \Rightarrow \frac{\sqrt{4n+3}\sqrt{4n+3}\sqrt{4n+7}}{(4n+5)\sqrt{4n+7}}\sqrt{\frac{3}{4n+3}}<&\sqrt{\frac{3}{4n+7}}\\ \Rightarrow \frac{\sqrt{4n+3}\sqrt{4n+7}}{(4n+5)}\sqrt{\frac{3}{4n+7}}<&\sqrt{\frac{3}{4n+7}}\\ \Rightarrow \frac{\sqrt{4n+3}\sqrt{4n+7}}{(4n+5)}<&1\\ \Rightarrow (4n+3)(4n+7)<&(4n+5)^2\\ \Rightarrow 16n^2+40n+21<&16n^2+40n+25\\ \Rightarrow 21<&25 \end{align*}

give us a true result for $n>0$.