If $0 < x < y$, show that $x < (x+y)/2 < y$ is true

Question

If $0 < x < y$, show that $x < (x+y)/2 < y$ is true.

I think this is an easy question, but I just can't see its solution. I thought I could add $x$ to all the terms, then I'd get $x < 2x < y + x$. From this, I could say that $2x < y +x$, so $x < (y+x)/2$. But I can't think in anything to keep with the solution.

Answer 1

Since $x<y,$ $$ 2x = x+x < x+y < y+y = 2y. $$ Now divide by $2$ throughout.

Answer 2

If $0<x<y$ then $y<x+y<2y$ implies $\frac{y}{2}<\frac{x+y}{2}<y$.

If $x<\frac{x+y}{2}$ then $2x=x+x<x+y$, which is true. Hence $x<\frac{x+y}{2}<y$, as required.

Answer 3

Alternatively, since $x<y$, you can consider $y=x+a, a>0$. Now substitute: $$\color{red}x < \color{green}{(x+y)/2} < \color{blue}y \iff \\ \color{red}x<\color{green}{\frac{x+x+a}{2}=x+\frac a2}<\color{blue}{x+a},$$ which is true.