The numbers x,y zx,yz satisfy abs(x+2)+abs(y+3)+abs(z-5)=1|x+2|+|y+3|+|z−5|=1 then prove that abs(x+y+z)<=1|x+y+z|≤1?
1 Answer
Jul 25, 2018
Please see Explanation.
Explanation:
Recall that,