Proving $\frac12\cdot\frac34\cdot\dots\cdot\frac{2n-1}{2n}\leq\frac1{\sqrt{3n+1}} ,\;\forall n\in\mathbb{N}$ using induction

Question

Base case. Let $n=1$, then $\frac12\leq\frac1{\sqrt{3+1}}$.

Induction step. Let's assume the inequality is true for some $k\in\mathbb{N}$.

We need to show that it's true for $k+1$, i.e. $\frac12\cdot\frac34\cdot\dots\cdot\frac{2k-1}{2k}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+4}}$.

From the assumption we get that $\frac12\cdot\frac34\cdot\dots\cdot\frac{2k-1}{2k}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+1}}\cdot\frac{2k+1}{2k+2}$.

So now I need to show that $\frac1{\sqrt{3k+1}}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+4}}$. How should I do this?

Answer 1

After squaring of the both sides you'll get $$(2k+1)^2(3k+4)\leq4(k+1)^2(3k+1)$$ or $$19k\leq20k$$

Answer 2

squaring both sides of your inequality (the last one) $$\frac{1}{3k+1}\cdot \frac{(2k+1)^2}{(2k+2)^2}\le \frac{1}{3k+4}$$ this is equivalent to $$(3k+4)(2k+1)^2\le (3k+1)(2k+2)^2$$ simplifying this we obtain: $$k>0$$ and this is true.

Answer 3

Taking squares it is equivalent to $(2k+1)^2(3k+4)\leq (3k+1)(2k+2)^2$, exapanding everything you get $12k^3+12k^2+3k+16k^2+16k+4\leq 12k^3+24k^2+12k+4k^2+8k+4$ which is equivalent to $19k\leq 20k$ which is trivially true since $k\geq 0$.

Answer 4

So now I need to show that $\frac1{\sqrt{3k+1}}\cdot\frac{2k+1}{2k+2}\leq\frac1{\sqrt{3k+4}}$. How should I do this?

Squaring and clearing denominators yields $(2k+1)^2 (3k+4) \leq (2k+2)^2 (3k+1)$. Expanding yields $12k^3+28k^2+19k+4 \leq 12k^3+28k^2+20k+4$. This simplifies to $19 k \leq 20k$, which is trivially true for positive $k$. Thus proven.