Which fraction is greater?

$\frac{15^{16} + 1}{15^{17} + 1} \boxed{?} \frac{15^{15}+1}{15^{16}+1}$ Cross Multiply! $15^{32} + 2\cdot15^{16} + 1 \boxed{ ? } 15^{32} + 15^{17} + 15^{15} + 1$ $2\cdot 15^{16} \boxed{ ? } 15^{17} + 15^{15}$ $2 \cdot 15 \boxed{ ? } 15^2 + 1$ $30 \boxed{ < } 226$

Let $a=15^{15}$ and consider the following:

$\begin{aligned} \dfrac{\dfrac{15^{15}+1}{15^{16}+1}}{\dfrac{15^{16}+1}{15^{17}+1}} & = \frac{(15^{15}+1)(15^{17}+1)}{(15^{16}+1)^2} = \frac{(a+1)(225a+1)}{(15a+1)^2} = \frac{225a^2+226a+1}{225a^2+30a+1} > 1 \end{aligned}$

$\Rightarrow \frac{15^{16}+1}{15^{17}+1} \boxed{<} \frac{15^{15}+1}{15^{16}+1}$

Let $\displaystyle A=\frac{15^{16}+1}{15^{17}+1}$ , $\displaystyle B=\frac{15^{15}+1}{15^{16}+1}$

Factor B:

$B=\frac{15^{15}+1}{15^{16}+1}=\frac{15\times(15^{15}+1) }{15\times(15^{16}+1)} =\frac{15^{16}+15}{15^{17}+15}=\frac{(15^{16}+1)+14}{(15^{17}+1)+14} >A$

So $\displaystyle \frac{15^{16}+1}{15^{17}+1}\boxed{<}\frac{15^{15}+1}{15^{16}+1}$

We simplify $A - B$ and look at the sign of the answer to determine which fraction is bigger: $A-B = \frac{15^{16}+1}{15^{17}+1} - \frac{15^{15}+1}{15^{16}+1} = \frac{(15^{16}+1)^2-(15^{15}+1)(15^{17}+1)}{(15^{17}+1)(15^{16}+1)}$ We simplify the numerator and get $(15^{16}+1)^2-(15^{15}+1)(15^{17}+1) \\ =15^{32}+2 \cdot 15^{16}+1-15^{32} -15^{15}-15^{17}-1 \\ = 2\cdot 15 \cdot 15^{15}-15^{15}-225 \cdot 15^{15} \\ = (-196) \cdot 15^{15}$ The numerator is negative, so $B$ must be greater than $A$ .

Multiply both numerator and denominator of B by 15. Then use the concept a/b < a+x/b+x

\[ We note that the denominators are larger than the numerators, so we compare 1/A and 1/B

1/A = 15 - 14/(15^16. +1) and. 1/B = 15- 14/(15^15 +1) Therefore 1/A > 1/B A < B \]

Assume that $B$ is larger and write the ratio $\frac{B}{A}$ as $\frac{(a^a + 1)(a^{a+2}+1)}{(a^{a+1} + 1)^2}$ where we set $a=15$ . Multiplying it out and gathering like terms yields $\frac {a^{2(a+1)} + a^a + a^{a+2} +1} {a^{2(a+1)} + 2a^{a+1} + 1 }.$ To check whether $B$ really is larger, this has to be bigger than $1$ , and since two terms in the numerator and the denominator are equal, it is sufficient to check that $a^a + a^{a+2}$ is larger than $2a^{a+1}$ . Again, with ratios, this becomes $\frac {a^a + a^{a+2}} {2a^{a+1}}= \frac{1+a^2}{2a} > 1 \quad\text{iff}\quad a>1.$

Hence, $B$ is larger than $A$ .

We know : With X,Y,Z >0 : X/Y < (X+Z)/(Y+Z) . With 0 < Z1 < Z2, There is Z3 > 0 that Z1 + Z3 = Z2 . ==> (X+Z1)/(Y+Z1) < (X+Z2)/(Y+Z2) with Z1 < Z2 .

A = (1 + 1/15^16)/(15 + 1/15^16) . And B = (1 + 1/15^15)/(15 + 1/15^15) . So A < B.

The larger fraction is the one whose denominator is small. So naturally B!