Note that for all integers $ 0 \le k \le n$, \[ x_n a^n x_k b^k \ge x_n b^n x_k a^k , \] with equality only when $ x_k= 0$ or $ k=n$. In particular, we have strict inequality when $ k=n-1$. In summation, this becomes \[ x_n a^n \sum_{k=0}^n x_k b^k > x_n b^n \sum_{k=0}^n x_k a^k, \] i.e., \[ x_n a^n \cdot B > x_n b^n \cdot A . \] Then \[ A'B = (A- x_n a^n)B = AB - x_n a^nB < AB - x_n b^n A = A(B- x_n b^n) = AB', \] as desired. $ \blacksquare$

Note that, $$a^{n}b^{n-1} >b^{n}a^{n-1}$$and hence $$a^{n}b^{n-1}x_{n-1} >b^{n}a^{n-1}x_{n-1}$$$$\Rightarrow a^{n}\left( b^{n-1}x_{n-1}+\ldots +b^{0}x_{0}\right) >b^{n}\left( a^{n-1}x_{n-1}+\ldots +a^{0}x_{0}\right)$$$$\Rightarrow a^{n}B_{n-1} >b^{n}A_{n-1}$$$$\Rightarrow \dfrac{a^{n}}{A_{n-1}} >\dfrac{b^{n}}{B_{n-1}}$$$$\Rightarrow \dfrac{x_{n}a^{n}}{A_{n-1}} >\dfrac{x_{n}b^{n}}{B_{n-1}}$$$$\Rightarrow \dfrac{A_{n-1}}{a^{n}x_{n}+A_{n-1}} <\dfrac{B_{n-1}}{b^{n}x_{n}+B_{n-1}}$$$$\Rightarrow \dfrac{A_{n-1}}{A_{n}} <\dfrac{B_{n-1}}{B_{n}} $$