Explanation:

we have

−

|

y

|

+

x

√

(

x

2

+

y

2

−

1

)

≥

1

⇒

x

−

|

y

|

≥

1

+

√

(

x

2

+

y

2

−

1

)

i

f

x

≥

|

y

|

,

then squaring both sides

x

2

y

2

−

2

x

|

y

|

≥

+

x

2

+

y

2

−

1

+

2

√

(

x

2

+

y

2

−

1

)

⇒

−

x

|

y

|

≥

√

(

x

2

+

y

2

−

1

)

.

.

.

.

.

.

.

.

.

.

.

.

(

1

)

since

x

≥

|

y

|

≥

0

then L.H.S. of (1) is non positive and R.H.S of (2) is non negative. Therefore the system is satisfied only when both sides are zero.

∴

The inequality (1) is equivalent to the system

{

x

|

y

|

=

0

x

2

+

y

2

−

1

=

0

THe first equation gives x=0 or y=0. if x=0, then we find

y

±

1

from second equation but

x

≥

|

y

|

which is impossible.

If y=0 then from second equation we find

x

2

=

1

∴

x=1,-1

taking x=1

(

∵

x

≥

|

y

|

)

∴

the pair (1,0) satisfies the given inequation. Hence (1,0) is the solution of the original

inequation.