Solve: x - 1 - | x | < 0

Given: x < 1 - | x | . . . ( 1 ) First we will define the term under the square root. 1 - | x | ≥ 0 ⇒ | x | ≤ 1 ⇒ - 1 ≤ x ≤ 1 ( If | x | ≤ a ( a ≥ 0 ) then - a ≤ x ≤ a ) x = 1 does not satisfy the above inequality. Now we take 2 cases: ( a ) - 1 ≤ x < 0 x < 1 + x as LHS is negative and RHS is positive, we know that negative is less than positive so, x ∈ [ - 1 , 0 ) . . . ( i ) ( b ) Now consider 0 ≤ x < 1 . . . ( a ) Since x < 1 - x squaring both sides, we get x 2 < 1 - x ⇒ x 2 + x - 1 < 0 ⇒ x + 1 2 2 < 5 4 ⇒ x + 1 2 < 5 2 Use modulus inequality - 5 2 < x + 1 2 < 5 2 ⇒ - 5 - 1 2 < x < 5 - 1 2 . . . ( b ) Taking ( a ) ∩ ( b ) , 0 ≤ x < 5 - 1 2 . . . ( ii ) So, combining ( i ) and ( ii ) , we get: x ∈ [ - 1 , 5 - 1 2 ) .

For a , b ∈ R , prove that | a + b | 1 + | a + b | ≤ | a | 1 + | a | + | b | 1 + | b | .

Given a , b ∈ R We have to prove that | a + b | 1 + | a + b | ≤ | a | 1 + | a | + | b | 1 + | b | We know that | a + b | ≤ | a | + | b | ⇒ 1 | a + b | ≥ 1 | a | + | b | 1 + 1 | a + b | ≥ 1 + 1 | a | + | b | 1 1 + 1 | a + b | ≤ 1 1 + 1 | a | + | b | | a + b | 1 + | a + b | ≤ 1 1 + 1 | a | + | b | . . . ( 1 ) and | a | 1 + | a | + | b | 1 + | b | ≥ | a | 1 + | a | + | b | + | b | 1 + | b | + | a | | a | 1 + | a | + | b | 1 + | b | ≥ | a | + | b | 1 + | a | + | b | | a | 1 + | a | + | b | 1 + | b | ≥ 1 1 + 1 | a | + | b | . . . ( 2 ) From equations ( 1 ) and ( 2 ) , we get: | a | + | b | 1 + | a | + | b | ≤ | a | 1 + | a | + | b | 1 + | b |