Finding the truth

Applying Cauchy-Schwarz inequality on sets $\big\{\binom{n}{0},\binom{n}{1},...,\binom{n}{n} \big\}$ & $\{1,1,...,1\}$ $\displaystyle\sum_{k=0}^n \binom{n}{k} \leq \sqrt{(1+1+...+1)} \sqrt{ {\binom{n}{0}}^2 + {\binom{n}{1}}^2 +...+ {\binom{n}{n}}^2 }$ $\Rightarrow 2^{n} \leq \sqrt{(n+1)} \sqrt {\binom{2n}{n}}$ $\Rightarrow 2^{2n} \leq (n+1) \binom{2n}{n}$

It is Choice A above, which can be proved via Induction:

CASE I ( $n=1$ ): $2^{2 \cdot 1} \le (1+1) \cdot {2 \choose 1} \Rightarrow 4 \le 4$ (TRUE).

CASE II ( $n=k$ ): $2^{2k} \le (k+1) \cdot {2k \choose k}$ (ASSUMED TRUE).

CASE III ( $n=k+1)$ : $2^{2k}2^{2} \le 2^{2} \cdot (k+1) \cdot {2k \choose k} \Rightarrow 2^{2(k+1)} \le 4(k+1) \cdot {2k \choose k};$

or $2^{2(k+1)} \le 4 \cdot (k+1)(\frac{k+2}{k+2}) \cdot {2k \choose k}\frac{{2(k+1) \choose k+1}}{{2(k+1) \choose k+1}};$

or $2^{2(k+1)} \le [((k+1)+1) \cdot {2(k+1) \choose k+1}] \cdot [\frac{4k+4}{k+2} \cdot \frac{(2k)!}{(k!)^2} \frac{((k+1)!)^2}{(2k+2)!}];$

or $2^{2(k+1)} \le [((k+1)+1) \cdot {2(k+1) \choose k+1}] \cdot [\frac{4k+4}{k+2} \cdot \frac{(2k)!}{(k!)^2} \frac{(k+1)^2 (k!)^2}{(2k+2)(2k+1)(2k)!}];$

or $2^{2(k+1)} \le [((k+1)+1) \cdot {2(k+1) \choose k+1}] \cdot [\frac{2(k+1)^2 }{(k+2)(2k+1)}];$

or $2^{2(k+1)} \le [((k+1)+1) \cdot {2(k+1) \choose k+1}] \cdot [\frac{2k^2 + 4k +2}{2k^2 + 5k +2}];$

or $2^{2(k+1)} \le [((k+1)+1) \cdot {2(k+1) \choose k+1}] \cdot [1 - \frac{k}{2k^2 + 5k +2}] \le [(k+1)+1] \cdot {2(k+1) \choose k+1};$

or $\boxed{2^{2(k+1)} \le [(k+1)+1] \cdot {2(k+1) \choose k+1}}.$ (TRUE).

$\mathbb{Q.}\mathbb{E.}\mathbb{D.}$