Inequality with powers of natural numbers

[Guest]

It is known that 2018²⁰¹⁹ is bigger than 2019²⁰¹⁸. Show that 2019²⁰²⁰ is greater than 2020²⁰¹⁹.

[shyamjayakannan]

[tex]\frac{{2018}^{2019}}{{2019}^{2018}}>1\Rightarrow\left(\frac{2018}{2019}\right)^{2018}2018>1\Rightarrow\left(\frac{2019}{2020}\times\frac{2018\times2020}{{2019}^2}\right)^{2018}>\frac{1}{2018}\Rightarrow\left(\frac{2019}{2020}\right)^{2018}>\frac{1}{2018}\left(\frac{{2019}^2}{2018\times2020}\right)^{2018}[/tex]

[tex]\Rightarrow\left(\frac{2019}{2020}\right)^{2019}>\frac{1}{2018}\left(\frac{{2019}^2}{2018\times2020}\right)^{2018}\frac{2019}{2020}\Rightarrow\left(\frac{2019}{2020}\right)^{2019}2019>\frac{2019}{2018}\left(\frac{{2019}^2}{2018\times2020}\right)^{2018}\frac{2019}{2020}[/tex]

[tex]\Rightarrow\frac{{2019}^{2020}}{{2020}^{2019}}>\left(\frac{{2019}^2}{2018\times2020}\right)^{2019}[/tex]. Now, the term on the RHS is > 1 because the inner fraction is > 1 and exponentiating it will still keep it > 1.

So, [tex]\Rightarrow\frac{{2019}^{2020}}{{2020}^{2019}}>1\Rightarrow\boxed{{2019}^{2020}>{2020}^{2019}}[/tex]