The numbers x,y zx,yz satisfy abs(x+2)+abs(y+3)+abs(z-5)=1|x+2|+|y+3|+|z5|=1 then prove that abs(x+y+z)<=1|x+y+z|1?

1 Answer
Jul 25, 2018

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.