Question

21 PUNCTE
demonstrati ca oricare ar fi a,b,,c∈ R are loc inegalitatea
$|2a+b-1| + | a-2b +3 | + |3a -b-5 | \geq 1$

Answer 1

1.Scoatem modulurile:
2a+b+1+a-2b+3+3a-b-5$\geq 1$
2.Rezlovam operatiile:
6a-2b-3$\geq 1$
3.Dam factor comun:
2(3a-b)-3