Answer:
Step-by-step explanation:
Given x,y,a&b are greater than zero
also [tex]\frac{x}{y}[/tex][tex]\leq [/tex][tex]\frac{a}{b}[/tex]
since x,y,a&b are greater than zero therefore we can cross multiply them without changing the inequality
therefore
[tex]\frac{x}{a}[/tex][tex]\leq [/tex][tex]\frac{y}{b}[/tex]
adding 1 on both sides we get
[tex]\frac{x}{a}[/tex]+1[tex]\leq [/tex][tex]\frac{y}{b}[/tex]+1
[tex]\frac{x+a}{a}[/tex][tex]\leq [/tex][tex]\frac{y+b}{b}[/tex]
rearranging
[tex]\frac{x+a}{y+b}[/tex][tex]\leq [/tex][tex]\frac{a}{b}[/tex]
