1. ## Prove this ...

a,b,x,y are a positif reals and x<y
Prove :
$\displaystyle \frac{x}{y}\prec(\frac{ax+by}{bx+ay}\prec\frac{y}{ x}$

2. Hello, dhiab!

$\displaystyle a,b,x,y$ are positive reals and $\displaystyle x<y.$

Prove: .$\displaystyle \frac{x}{y} \:<\:\frac{ax+by}{bx+ay} \;<\:\frac{y}{x}$

$\displaystyle \begin{array}{ccccc}\text{We have:} & x \:<\:y \\ \text{Square:} & x^2 \:<\:y^2 \\ \text{Multiply by }b\!: & bx^2 \:<\:by^2 \\ \text{Add }axy\!: & bx^2 + axy \:<\:axy + by^2 \\ \text{Factor:} & x(bx+ay) \:<\:y(ac+by) \\ \text{Hence:} & \qquad\;\; \dfrac{x}{y} \:<\:\dfrac{ax+by}{bx+ay} \end{array}$

$\displaystyle \begin{array}{ccccc}\text{We have:} & x \:<\:y \\ \text{Square:} & x^2 \:<\:y^2 \\ \text{Multiply by }a\!:& ax^2 \:<\:ay^2 \\ \text{Add }bxy\!: & ax^2 + bxy \:<\:bxy + ay^2 \\ \text{Factor:} & x(ax+by) \:<\:y(bx+ay) \\ \text{Hence:} & \dfrac{ax+by}{bx+ay} \:<\:\dfrac{y}{x} \qquad\;\; \end{array}$

. . . $\displaystyle \text{Therefore: }\;\frac{x}{y} \;<\;\frac{ax+by}{bx + ay} \;<\:\frac{y}{x}$