Triangle inequlity improvment with the angle conditions

Question

I was working on how to proof $a+b \leq x+y+z$?

Apply triangle inequity to the triangle ADC,

$x+z \geq a$

Apply triangle inequity to the triangle DCB,

$y+b \geq z$

Adding above inequities,

$x+y+z+b \geq a+z$

I am getting

$x+y \geq a-b$ , But how do I proof

$a+b \leq x+y+z$?

Note that the angles with the horizontal line $AC < 10^\circ$ , $CB < 10^\circ$ $AB < 10^\circ$ and $DB < 10^\circ$

Answer 1

The claim is simply not true in general. You see that $x+z\approx a$ and also $y+b\approx z$ if the angles are all so small. But then $a+b\approx x+2z-y$ and that is definitely greater than $x+y+z$ as in your sketch $z>2y$.

Answer 2

Fix $A, B, C$, let $D$ move along $AC$ from $C$ to $A$. By your assumption $BD$ is becoming shorter, so $\min z= b$. It remains to show $AC\ge AB$, this is WRONG! So your inequality is not correct in genaral.