Use proof by induction to show that for any positive integer $n\geq 2$ the following holds

Question

Use proof by induction to show that for any positive integer $n\geq 2$ the following holds:

$(1+\frac{1}{3})(1+\frac{1}{5})...(1+\frac{1}{2n-1})>\frac{\sqrt{2n+1}}{3}$

Proof: Base case: $n=2$. $LHS:\frac{4}{3}$; $RHS:\frac{\sqrt{5}}{3}$. Clearly, $\frac{4}{3}>\frac{\sqrt{5}}{3}$.

Next, we assume that for some positive integer $n\geq2$ the above statement holds.

Here is where I get consuded. Do I add the next term i.e $(1+\frac{1}{2n})$ to both sides? Or do I multiply the right hand side by that term to continue? Any tips on how to tackle induction proofs like these? Thanks!

Answer 1

Assuming that $(1+\frac{1}{3})(1+\frac{1}{5})...(1+\frac{1}{2n-1})>\frac{\sqrt{2n+1}}{3}$ consider $(1+\frac{1}{3})(1+\frac{1}{5})...(1+\frac{1}{2n-1})(1+\frac1 {2n+1})$. This quantity is greater than $\frac{\sqrt{2n+1}}{3} (1+\frac1 {2n+1})$. We want to show that $\frac{\sqrt{2n+1}}{3} (1+\frac1 {2n+1}) >\frac{\sqrt{2n+3}}{3}$. For this use the fact that $\sqrt{2n+1} (1+\frac1 {2n+1})=\sqrt {2n+1} +\frac 1 {\sqrt {2n+1}}$. Puting $x=2n+1$ you will now require the inequality $\sqrt x +\frac 1 {\sqrt x } >\sqrt {x+2}$ or $x+1 >\sqrt {x(x+2)}$. But $x(x+2)=x^{2}+2x < x^{2}+2x+1=(x+1)^{2}$ and you can finish the proof by taking square root on both sides.

Answer 2

The base you checked.

By the assumption of the indiction $$\prod_{k=1}^{n+1}\left(1+\frac{1}{2k-1}\right)>\frac{\sqrt{2n+1}}{3}\left(1+\frac{1}{2n+1}\right).$$ Thus, it's enough to prove that: $$\frac{\sqrt{2n+1}}{3}\left(1+\frac{1}{2n+1}\right)>\frac{\sqrt{2n+3}}{3}$$ or $$2n+2>\sqrt{(2n+3)(2n+1)}$$ or $$2n+2>\sqrt{(2n+2)^2-1},$$ which is obvious.