Induction, combinatorics, inequality proof $\binom{2n}{n} < 2^{2n - 2}$ for all $n \geq 5$

Show that $$\binom{2n}{n} < 2^{2n - 2}$$ for all $$n \geq 5$$.

Can I get some help please? I started by trying to prove this by induction, For the base case $$n=5$$, I showed the inequality is true. For the the induction step, I assumed that the inequality is true and started trying to show it's true for $$n+1$$. So $$\binom{2n+2}{n+1}$$ and I broke it down by combinatorics to get $$2\binom{2n}{n}+\binom{2n}{n+1}+\binom{2n}{n-1}$$ which is less than $$2\cdot 2^{2n-2}+\binom{2n}{n+1}+\binom{2n}{n-1}$$ using the inductive step and I don't know how to continue.

• Use factorial formula for ${2n+2 \choose n+1}$ and ${2n \choose n}$, then use $\frac{(2n+1)(2n+2)}{(n+1)(n+1)} < 4$. Ratio $\dfrac{2n+2 \choose n+1}{2n \choose n}=\frac{(2n+1)(2n+2)}{(n+1)(n+1)}$ also can be shown using combinatorics. Apr 15, 2022 at 10:54
• Please type your question rather than posting an image since images cannot be searched. Apr 15, 2022 at 11:50

Base Case: $$n = 5$$ can be checked manually.

Now, we need to show that $$\binom{2n + 2}{n + 1} < 2 ^ {2n}$$ assuming that $$\binom{2n}{n} < 2 ^ {2n - 2}$$

Note that it suffices to show that $$\frac{\binom{2n + 2}{n + 1}}{\binom{2n}{n}} < 4 \text{ (Why?)}$$

Expanding and cross-multiplying (all terms are positive), we get $$\frac{(2n + 1)(2n + 2)}{(n + 1)(n + 1)} < 4$$ $$\iff \frac{2n + 1}{n + 1} < 2$$ which is true.

Proof:

When $$n=5$$, we have $$252=\left( \begin{array}{c}10\\5\\\end{array} \right) \le 4^4=256$$.

Then $$\forall n\ge 5$$, suppose that $$\left( \begin{array}{c}2n\\n\\\end{array} \right) < 4^{n-1}$$ is true, and we have:

\begin{align} &1<2\\ \Rightarrow ~~&2n+1<2n+2\\ \Rightarrow ~~&\frac{2n+1}{n+1}<2\\ \Rightarrow ~~&\frac{\left( 2n+1 \right) \left( 2n+2 \right)}{\left( n+1 \right) \left( n+1 \right)}<4\\ \Rightarrow ~~&\frac{\left( 2n+1 \right) \left( 2n+2 \right)}{\left( n+1 \right) \left( n+1 \right)}\times \left( \begin{array}{c} 2n\\ n\\ \end{array} \right) <4^n\\ \Rightarrow ~~&\frac{\left[ 2\left( n+1 \right) \right] !}{\left( n+1 \right) !\left( n+1 \right) !}<2^{2\left( n+1 \right) -2}, \end{align}

i.e., the original statement is also true for $$(n+1)$$.

So $$\left( \begin{array}{c}2n\\n\\\end{array} \right) < 2^{2n-2}$$ holds for all $$n\ge 5$$.

P.S.

I see that @Ivan Kaznacheyeu has given the tip, so this answer is just an implementation of it.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0}