Prove that if $x<y$, then $3x+2y<2x+3y<5y$

Question

I am stuck with the following problem:

Let $x,y \in \mathbb{R}$, prove that if $x<y$, then $$3x+2y<2x+3y<5y$$

I have the following:

Suppose $x<y$, then

$x+2x<y+2x$

$ \Rightarrow $ $3x<y+2x$

$ \Rightarrow $ $3x+2y<y+2x+2y$

$ \Rightarrow $ $3x+2y<3y+2x$

$ \Rightarrow $ $3x+2y<2x+3y$

How can I further prove that $3x+2y<2x+3y<5y$ ?

Answer 1

Start from $x<y$. Add $2x$ to both sides to get $3x<y+2x$. Now add $2y$ to both sides to get $3x+2y<3y+2x$.

As for the second inequality it is even easier. If $x<y$ then $2x+3y<2y+3y=5y$.

Answer 2

We have $$2x+3y<2y+3y = 5y$$

and $$2x+3y= 2x+2y+y>2x+2y+x = 3x+2y$$

Answer 3

You can prove that $3x+2y<2x+3y$ observing that$$(2x+3y)-(3x+2y)=y-x>0.$$And, since $x<y$, $2x+3y<2y+3y=5y$.