Prove y is less than or equal to z

[anemone]

Let $x,\,y,\,z,\,a,\,b$ be positive real such that $xa+yb\le ya+zb\le za+xb$.

Prove that $y\le z$.

 

[Albert1]

Let $x,\,y,\,z,\,a,\,b$ be positive real such that $xa+yb\le ya+zb\le za+xb$.

Prove that $y\le z$.

Spoiler

if $y>z------(1)$
we have:$ay+bz\leq az+bx<ay+bx$
$\therefore bz<bx,\, or \,,x>z----(2)$
$az+bx\geq ax+by>az+by$
$\therefore bx>by ,\,\, or \,\, x>y---(3)$
from (1)(2)(3)we get :$x>y>z$
this contradicts to $ax+by\leq ay+bz$
so we obtain $y\leq z$
it is easy to see $y=z=x$ is possible

 

[anemone]

Thanks for participating, Albert!

Solution of other:

Spoiler

The first inequality tells us:

$$\color{yellow}\bbox[5px,purple]{xa+yb\le ya+zb}\color{black}\le za+xb$$

$$(y-z)b\le a(y-x)$$

Whereas the second inequality is equivalent to

$$\color{black}xa+yb\le\color{yellow}\bbox[5px,green]{ ya+zb\le za+xb}$$

$$(y-z)a\le b(x-z)$$

If $y>z$, then the left sides of the two inequalities above are positive, so the right sides are positive as well.

In particular, this means $y-x>0$ and $x-z>0$, this further implies $y>x>z$, giving

$xa>za$ and $yb>xb$

Adding them up gives $xa+yb>za+xb$, which contradicts the relation between the first and third expressions in the given original inequality.

Thus, $y\le z$.