Comparing $\left({\frac{2^{10}}{11}}\right)^{11}$ and $\binom{10}1^2\cdot\binom{10}2^2\cdot\binom{10}3^2\cdot\binom{10}4^2\cdot\binom{10}5$

Question

is

A. Strictly larger than $\binom{10}{1}^2 \cdot \binom{10}{2}^2 \cdot \binom{10}{3}^2 \cdot \binom{10}{4}^2 \cdot \binom{10}{5}$

B. Strictly larger than $\binom{10}{1}^2 \cdot \binom{10}{2}^2 \cdot \binom{10}{3}^2 \cdot \binom{10}{4}^2$ but strictly smaller than $\binom{10}{1}^2 \cdot \binom{10}{2}^2 \cdot \binom{10}{3}^2 \cdot \binom{10}{4}^2 \cdot \binom{10}{5}$

C. less than or equal to $\binom{10}{1}^2 \cdot \binom{10}{2}^2 \cdot \binom{10}{3}^2 \cdot \binom{10}{4}^2$

D. Equal to $\binom{10}{1}^2 \cdot \binom{10}{2}^2 \cdot \binom{10}{3}^2 \cdot \binom{10}{4}^2 \cdot \binom{10}{5}$

Here is the answer I already wrote that very day here:

Indeed using AM-GM inequality :

$(\frac{2^{10}}{11})^{11} = (\frac{1+\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4}+\binom{10}{5}+\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}}{11})^{11}$

=$(\frac{2+2 \cdot (\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4})+\binom{10}{5}}{11})^{11}$

Now we know that $(\frac{2+2 \cdot (\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4})+\binom{10}{5}}{11}) > [\binom{10}{1}^{2}\cdot \binom{10}{2}^{2} \cdot \binom{10}{3}^{2} \cdot \binom{10}{4}^{2} \cdot \binom{10}{5} ]^{\frac{1}{11}}$.... [By AM-GM Inequality]

Also $(\frac{2+2 \cdot (\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4})+\binom{10}{5}}{11})^{11}= (\frac{2^{10}}{11})^{11}$

$\implies (\frac{2^{10}}{11})^{11} > \binom{10}{1}^{2}\cdot \binom{10}{2}^{2} \cdot \binom{10}{3}^{2} \cdot \binom{10}{4}^{2} \cdot \binom{10}{5}$

Hence option A (proved)

I did this all by myself. No plagiarism. No help sought.No books used and no internet reference used. Last time I answered I was alleged to have plagiarized here on this platform.

Answer 1

Indeed using AM-GM inequality :

$(\frac{2^{10}}{11})^{11} = (\frac{1+\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4}+\binom{10}{5}+\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}}{11})^{11}$

=$(\frac{2+2 \cdot (\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4})+\binom{10}{5}}{11})^{11}$

Now we know that $(\frac{2+2 \cdot (\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4})+\binom{10}{5}}{11}) > [\binom{10}{1}^{2}\cdot \binom{10}{2}^{2} \cdot \binom{10}{3}^{2} \cdot \binom{10}{4}^{2} \cdot \binom{10}{5} ]^{\frac{1}{11}}$.... [By AM-GM Inequality]

Also $(\frac{2+2 \cdot (\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4})+\binom{10}{5}}{11})^{11}= (\frac{2^{10}}{11})^{11}$

$\implies (\frac{2^{10}}{11})^{11} > \binom{10}{1}^{2}\cdot \binom{10}{2}^{2} \cdot \binom{10}{3}^{2} \cdot \binom{10}{4}^{2} \cdot \binom{10}{5}$

Hence option A (proved)

I did this all by myself. No plagiarism. No help sought.No books used and no internet reference used. Last time I answered I was alleged to have plagiarized here on this platform.