Question

The numbers `x,y z` satisfy `abs(x+2)+abs(y+3)+abs(z-5)=1` then prove that `abs(x+y+z)<=1`?

Answer 1

Please see Explanation.

Explanation:

Recall that, `|(a+b)| le |a|+|b|............(star)`.

`:. |x+y+z|=|(x+2)+(y+3)+(z-5)|`,

`le |(x+2)|+|(y+3)|+|(z-5)|....[because, (star)]`,

`=1...........[because," Given]"`.

`i.e., |(x+y+z)| le 1`.