Integral Exercise asked by Dan Sitaru in many math groups

Exercise:

Show that, $11\int_{0}^{1}{\prod\limits_{i=1}^{10}{\left( x+i \right)}\,dx}<{{6}^{11}}$

Solution: we know that there is a relation between athematic mean and geometric mean

So $AM\ge GM\Leftrightarrow \frac{\sum\limits_{i=1}^{10}{\left( x+i \right)}}{10}\ge \sqrt[10]{\prod\limits_{i=1}^{10}{\left( x+i \right)}}$

But $\sum\limits_{i=1}^{10}{\left( x+i \right)=\sum\limits_{i=1}^{10}{x}+\sum\limits_{i=1}^{10}{i}=10x+\sum\limits_{i=1}^{10}{i}=10x+\frac{10\times 11}{2}=10x+55}$

So $\frac{10x+55}{10}\ge \sqrt[10]{\prod\limits_{i=1}^{10}{\left( x+i \right)}}\Leftrightarrow {{\left( x+\frac{55}{10} \right)}^{10}}\ge \prod\limits_{i=1}^{10}{\left( x+i \right)}$

$\Leftrightarrow \int_{0}^{1}{{{\left( x+\frac{55}{10} \right)}^{10}}dx\ge \int_{0}^{1}{\prod\limits_{i=1}^{10}{\left( x+i \right)dx}}}$

Let $u=x+\frac{55}{10}\Leftrightarrow du=dx$

So $\int_{0}^{1}{{{\left( x+\frac{55}{10} \right)}^{10}}dx}=\int_{\frac{11}{2}}^{\frac{13}{2}}{{{u}^{10}}du}=\frac{1}{11}\left( {{u}^{11}} \right)_{\frac{11}{2}}^{\frac{13}{2}}=\frac{1}{11}\left( {{\left( \frac{13}{2} \right)}^{11}}-{{\left( \frac{11}{2} \right)}^{11}} \right)=\frac{1}{11}\left( \frac{{{13}^{11}}-{{11}^{11}}}{{{2}^{11}}} \right)$

So $11\int_{0}^{1}{\prod\limits_{i=1}^{10}{\left( x+i \right)dx}\le \frac{{{13}^{11}}-{{11}^{11}}}{{{2}^{11}}}}$

Observe that, $12>2\Leftrightarrow {{12}^{11}}>{{2}^{11}}\Leftrightarrow {{12}^{11}}>{{\left( 13-11 \right)}^{11}}$

But ${{\left( 13-11 \right)}^{11}}<{{13}^{11}}-{{11}^{11}}$ so ${{12}^{11}}<{{13}^{11}}-{{11}^{11}}$

Also ${{12}^{11}}={{\left( 2\times 6 \right)}^{11}}={{2}^{11}}\times {{6}^{11}}<{{13}^{11}}-{{11}^{11}}\Leftrightarrow {{6}^{11}}<\frac{{{13}^{11}}-{{11}^{11}}}{{{2}^{11}}}$

Thus $11\int_{0}^{1}{\prod\limits_{i=1}^{10}{\left( x+i \right)dx<{{6}^{11}}}}$              Q.E.D





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The idea of solution credit to Diego Alvariz