Comparison of value between two fractions

[anemone]

Prove $\dfrac{3^{371}+5}{5^{247}+3}>\dfrac{2^{371}+3}{3^{247}+2}$.

 

[kaliprasad]

Prove $\dfrac{3^{371}+5}{5^{247}+3}>\dfrac{2^{371}+3}{3^{247}+2}$.

Spoiler

we have

$3^3 \gt 5^2$

or $(3^3)^ {123} * 9 \gt (5^2)^{123} * 5$

or $3^{371} \gt 5^{247}$

hence

$3^{371} + 5 \gt 5^{247} + 3$

or

$\dfrac{3^{371} + 5}{5^{247} + 3} \gt 1 \cdots(1)$ also

we have

$3^2 = 2^3 + 1$

or $(3^2)^ {120} = (2^3+1)^{120}$

or $3^{240} \gt 2^{360} + 120* (2^3)^{119}$ as $(x+1)^n \gt x^n + nx^{n-1}$ for x positive

or $3^{240} \gt 2^{360} + 1\cdots(2)$ we do not need more than one

further $3^7(= 2181) \gt 2^{11}(=2048)\cdots(3)$

from (2) and (3)

$3^{247} \gt 2^{371} + 2^{11}$

or $3^{247} + 2 \gt 2^{371} + 2^{11}+ 2$

or $3^{247} + 2 \gt 2^{371} + 3$ as $2^{11}+1 \gt 3$

or

$\dfrac{2^{371} + 3}{3^{247} + 2} \lt 1 \cdots(4)$



from (1) and (4) we get the result

 

[anemone]

Spoiler

we have

$3^3 \gt 5^2$

or $(3^3)^ {123} * 9 \gt (5^2)^{123} * 5$

or $3^{371} \gt 5^{247}$

hence

$3^{371} + 5 \gt 5^{247} + 3$

or

$\dfrac{3^{371} + 5}{5^{247} + 3} \gt 1 \cdots(1)$ also

we have

$3^2 = 2^3 + 1$

or $(3^2)^ {120} = (2^3+1)^{120}$

or $3^{240} \gt 2^{360} + 120* (2^3)^{119}$ as $(x+1)^n \gt x^n + nx^{n-1}$ for x positive

or $3^{240} \gt 2^{360} + 1\cdots(2)$ we do not need more than one

further $3^7(= 2181) \gt 2^{11}(=2048)\cdots(3)$

from (2) and (3)

$3^{247} \gt 2^{371} + 2^{11}$

or $3^{247} + 2 \gt 2^{371} + 2^{11}+ 2$

or $3^{247} + 2 \gt 2^{371} + 3$ as $2^{11}+1 \gt 3$

or

$\dfrac{2^{371} + 3}{3^{247} + 2} \lt 1 \cdots(4)$



from (1) and (4) we get the result

Very well done, kaliprasad!(Clapping)