denote the integer part, and respectively the fractional part. Proof. In inequality a a + 2b + 2c. + b. 2a + b + 2c. + c. 2a + 2b + c. â¥. 3. 5. , we take a = x, b = [x], ...
Inequalities for Integer and Fractional Parts Mihály Bencze Department of Mathematics, Áprily Lajos College, Brasov, Romania Florentin Smarandache Chair of Math & Science Dept., University of New Mexico, Gallup, NM 87301, USA
Abstract: In this paper we present some new inequalities relative to integer and functional parts. [x] {x} 4 + ≥ , where [] ⋅ and {} ⋅ 3x + {x} 3x + [x] 15 denote the integer part, and respectively the fractional part. b c 3 a Proof. In inequality + + ≥ , we take a + 2b + 2c 2a + b + 2c 2a + 2b + c 5 a = x, b = [x], c = {x} . Theorem 1.
If x > 0 , then
Theorem 2. If a, b, c, x > 0 , then b c 3 a + + ≥ . [x]b + {x}c [x]c + {x}a [x]a + {x}b x b c 3 a Proof. In inequality + + ≥ , we take u = [x] and ub + vc uc + va ua + vb u + v v = {x} . Theorem 3. If x > 0 and a ≥ 1 , then [x] 2a + 1 [x] + ≤ . (a + 1)[x] + 2{x} (a + 1){x} + 2[x] (a + 1)(a + 2) Proof. In inequality
y z 3 x + + ≤ , we take ax + y + z x + ay + z x + y + az a + 2
y = [x] and z = {x} . Theorem 4. If x > 0 , then
1
⎛ ⎞ ⎛ ⎞ 1 1 1 1 [ x] ⎜ + + ⎟ + {x} ⎜ ⎟ ≤1. ⎝ x[ x] + x + 1 [ x]{x} + [ x] + 1 ⎠ ⎝ x[ x] + x + 1 x{x} + {x} + 1 ⎠ y z x Proof. In inequality + + ≤ 1 , we take y = [x] xy + x + 1 yz + y + 1 zx + z + 1 and z = {x} . Theorem 5. If x > 0 , then x[x]2 x{x}2 3 x3 + + ≥ 2 2 2 2 2 2 2 [x] + 3[x]{x} + 3{x} [x] 3[x] + 3[x]{x} + {x} {x} [x] + [x]{x} + {x}
(
)
Proof. In inequality
(
∑ y (x
2
)
3 x2 ≥ , we take y = [x] and 2 x+ y+z + xy + y
)
z = {x} . 1 1 1 + ≥ . [x] + 2{x} 2[x] + {x} x a 2 + bc Proof. In inequality ∑ ≥ a + b + c , we take a = x, b = [x], c = {x} . b+c Theorem 6. If x > 0 ,
Theorem 7. If x > 0 ,
(
)
x 3[x]2 − {x}2 {x} 3 [x]3 + ≥ [x]2 + [x]{x} + {x}2 3{x}2 + 3[x]{x} + [x]2 3 3[x]2 + 3[x]{x} + {x}2
(
Proof. In inequality
c = {x} .
∑a
)
a a+b+c ≥ , we take a = x, b = [x] , 2 3 + ab + b 3
2
Theorem 8. If x > 0 , then 1 1 + 3 + 3 2 2 3 2 2[x] + 4[x] {x} + 4[x]{x} + {x} [x] + [x] {x} + [x]{x}2 + {x} 3 1 1 + 3 ≤ . 2 2 3 [x] + 4[x] {x} + 4[x]{x} + 2{x} x[x]{x} 1 1 Proof. In inequality ∑ 3 ≤ , we take a = x, b = [x], c = {x} . 3 a + b + abc abc
⎛ [x]3 {x} 3 ⎞ ≥ [x]2 + [x]{x} + {x}2 . + Theorem 9. If x > 1 , then 4 ⎜ ⎟ ⎝ {x} [x] ⎠ 1 3 Proof. In inequality ∑ (−a + b + c ) ≥ a 2 + b 2 + c 2 , we take a = x, b = [x], a c = {x} .
2
Theorem 10. If x > 0 , then
(
)
x [x]3 + {x} 3 3 x4 + ≥ x 2 + [x]{x} 2 2 2 2 2 [x] − [x]{x} + {x} [x] + [x]{x} + {x}
Proof. In inequality
c = {x} .
3∑ ab a3 ∑ b2 − bc + c2 ≥ a , we take ∑
a−b
∑a+b
∑
In
{x} > 1. x + [x]
[x] + x + {x}
x > 2 , we take y = [x], z = {x} . y+z
Theorem 13. If x > 1 , then 3 + Proof.
a = x, b = [x],
< 1 we take a = x, b = [x], c = {x} .
Theorem 12. If x > 0 , then Proof. In inequality
)
[x]{x} ([x] − {x} ) < 1. x (x + [x])(x + {x} )
Theorem 11. If x > 0 , then Proof. In inequality
(
{x} [x] + ≥3 [x] {x}
3
(x + [x])(x + {x}) . [x]{x}
⎛ ⎛ 1⎞ ≥ 3 a ∑ ⎜⎝ ∑ a ⎟⎠ ⎜⎜ 1 + ⎝
( )
inequality
3
∏ (a + b ) ⎞⎟ , ⎟⎠
abc
we
a = x, b = [x], c = {x} . 4
⎞ ⎛ [x]{x} Theorem 14. If x > 0 , then ⎜ [x] + + {x} ⎟ ≥ 32[x]{x} . x ⎠ ⎝
Proof. In inequality
∑
xy ≥ 2
4
xyz ∑ x , we take y = [x], z = {x} .
(
Theorem 15. If x > 0 , then x 2 + [x]{x} Proof. In inequality
) ≥ 6x [x]{x} . 2
2
(∑ xy) ≥ 3xyz∑ x , we take y = [x], 2
Theorem 16. If x > 0 , then x 2 − x [x]{x} + [x]{x} ≥ [x] {x} + {x} [x]
(
Proof. In inequality
∑ xy ≥ ∑ x
Theorem 17. If x > 0 , then
z = {x} .
) x.
yz , we take y = [x], z = {x} .
(
)
[x](x + {x} ) + {x} (x + [x]) ≤ 2 2 − 1 x .
3
take
Proof. In inequality
∑ x (y + z ) ≤
2 ∑ x , we take y = [x], z = {x} .
[x] {x} 1 + ≥ . x + {x} x + [x] 2 a 3 Proof. In inequality ∑ ≥ , we take a = x, b = [x], c = {x} . b+c 2 Theorem 18. If x > 0 , then
Theorem 19. If x > 0 , then (x + [x]) + (x + {x} ) ≥ 21x[x]{x} + [x]3 + {x} 3 . 3
Proof. In inequality
∑ (x + y )
3
Theorem 20. If x > 1 , then Proof. In inequality
x+y + x+z
3
≥ 21xyz + ∑ x 3 , we take y = [x], z = {x} .
x + [x] + x + {x}
x + {x} [x] {x} ≤ + . x + [x] {x} [x]
x+z y+z , we take y = [x], z = {x} . ≤ x+y yz
x x 5 + ≥ . x + [x] x + {x} 2 1 9 , we take y = [x], z = {x} . Proof. In inequality 2∑ ≥ x + y ∑x Theorem 21. If x > 0 , then
2
2
{x} ⎛ {x} ⎞ ⎛ [x] ⎞ {x}2 ⎛ 1 1 ⎞ Theorem 22. If x > 1 , then +⎜ +⎜ + 2 ≥⎜ + [x] . ⎟ ⎟ ⎝ x {x} ⎟⎠ [x] ⎝ [x] ⎠ ⎝ {x} ⎠ x x2 x Proof. In inequality ∑ 2 ≥ ∑ , we take y = [x], z = {x} . y z
Theorem 23. If x > 0 , then 3 2 [ x]2 − [ x]{x} + {x}2 ≥ max [ x]2 ; ([ x] − {x}) ;{x}2 . 4 3 2 2 2 Proof. In inequality ∑ x 2 − ∑ xy ≥ max ( x − y ) ; ( y − z ) ; ( z − x ) , we take 4 y = [x], z = {x} .
{
}
{
}
Theorem 24. If x > 0 , then e{ x} + e[ x ] ≥ 2 + x . Proof. In inequality e y + e z ≥ 2 + y + z , we take y = [x], z = {x} . Theorem 25. If x ∈R , then sin[x] + sin{x} + cos x ≥ 1 . Proof. In inequality sin a + sin b + cos(a + b) ≥ 1 we take a = x, b = {x} .
4
Theorem 26. If x > 0 , then ( 3[ x]2 + 3[ x]{x} + {x}2 ) ⋅ ([ x]2 + [ x]{x} + {x}2 ) ⋅
(
) (
)
3
⋅ 3{ x}2 + 3{ x}[ x ] + [ x ]2 ≥ x 2 + [ x ]{ x} .
Proof. In inequality 3 a 2 + ab + b 2 ⋅ b 2 + bc + c 2 ⋅ c 2 + ca + a 2 ≥ (ab + bc + ca ) , we take
(
)(
)(
)
a = x, b = [x], c = {x} . Theorem 27. If x > 0 , then [ x] {x} 11 . + ≥ ( 3[ x] + 2{x})([ x] + 2{x}) ( 3{x} + 2[ x])({x} + 2[ x]) 48 x Proof. In inequality ( ∑ x ) ∑ ( 2 x + y +xz )( y + z ) ≥ 98 , y = [x], z = {x} . Theorem 28. If x > 0 , then Proof. In inequality
(
take
)
x 2x 2 + 3[x]{x} [x]2 {x}2 + ≥ . x + [x] x + {x} (x + [x])(x + {x} )
x2 3 ∑ (x + y )(x + z ) ≥ 4 , we take y = [x], z = {x} .
Theorem 29. If x > 1 , then ⎛ 2{x} 2[x] [x] 1+ + 1+ ≥ 1+ 2⎜ + [x] {x} ⎝ x + {x}
Proof. In inequality
we
∑
{x} ⎞ . x + [x] ⎟⎠
y+z x ≤ 2∑ we take y = [x], z = {x} . x y+z
⎛ [x] {x} [x] {x} ⎞ + ≥ 1+ 2⎜ + . ⎝ x + {x} x + [x] ⎟⎠ [x] {x} y+z x ≥ 4∑ , we take y = [x], z = {x} . Proof. In inequality ∑ x y+z
Theorem 30. If x > 1 , then
Theorem 31. If x > 0 , then 1). min 2). Proof. In c = {x} , etc.
(
{(
)
2 +1
)
2 +1
x + {x };
(
)
2 +1
}
x + x + [ x] ≥ 5 ([ x] + 2{x})
x + x + {x } ≥ 5 ({x} + 2[ x]) .
a + b + c + b + c + c ≥ a + 4b + 9c , we take
5
a = x, b = [x],
Theorem 32. If x ∈R , then 1). sin x ≤ sin[x] + sin{x} 2). cos x ≤ cos[x] + cos{x} Proof.
In
sin (a + b ) ≤ sin a + sin b
inequalities
and
cos (a + b ) ≤ cos a + cosb , we take a = x, b = [x] . 3
{x} [x] ⎛ x [x] 3 {x} ⎞ Theorem 33. If x > 1 , then 6 + + ≥⎜3 +3 + . [x] {x} ⎝ [x] {x} x ⎟⎠
Proof.
In inequality 3
⎛
(∑ a )⎛⎜⎝ ∑ 1a ⎞⎟⎠ ≥ ⎜⎝ ∑
3
a⎞ , we take a = x, b = [x], b ⎟⎠
c = {x} . Theorem 34. If x > 0 , then Proof. In inequality
[x]
(x + {x})
2
a
∑ (b + c )
2
≥
9
+
4∑ a
{x}
(x + [x])
2
≥
1 . 8x
, we take a = x, b = [x], c = {x} .
Theorem 35. If x > 0 , then { x } (x + { x } ) x[ x ] [ x ] + 5{ x} . + ≥ 12 x (x + {x})(2 x + [ x ]) (x + [ x ])(2 x + {x}) 3 a(a + b) Proof. In inequality ∑ ≥ , we take a = x, b = [x], (b + c )(2a + b + c ) 4 c = {x} . [ x] { x} 3[ x ]2 + 4[ x ]{ x} + 3{ x}2 + ≤ . 2 x + { x} 2 x + [ x ] 6x ab 1 Proof. In inequality ∑ ≤ ∑ a , we take a = x, b = [x], c = {x} . a + b + 2c 4
Theorem 36. If x > 0 , then
Theorem 37. If x > 0 , then ([ x]5 − [ x]2 + 3)({x}5 − {x}2 + 3) ≥ Proof.
In
inequality
∏(a
5
− a 2 + 3) ≥ ( ∑ a ) ,
8 x3 . x5 − x 2 + 3
3
we
a = x, b = [x], c = {x} . Theorem 38. If x > 0 , then
(2x + [x])2 + (2x + {x})2 2 2 2[x]2 + (x + {x} ) 2{x}2 + (x + [x])
6
≤5.
take
( 2a + b + c ) ∑ 2a 2 + b + c 2 ≤ 8 , we take ( ) 2
Proof. In inequality
Theorem 39. If x > 0 , then
( 2). ([ x] + 3). ({x} +
)
a = x, b = [x], c = {x} .
1). x + x[ x] + 3 x[ x]{x} ≤ 9 x 2 ( x + [ x]) 3
)
3
[ x]{x} + 3 x[ x]{x} ≤ 9 x 2 [ x]
)
x{x} + 3 x[ x]{x} ≤ 9 x 2 ( x + {x}) .
Proof. In inequality
3
a + ab + 3 abx ⎛ a + b⎞ ⎛ a + b + c⎞ ≤ 3 a⎜ ⎟⎠ , we take ⎝ 2 ⎟⎠ ⎜⎝ b 2
a = x, b = [x], c = {x} , etc.
(
)
Theorem 40. If x > 0 , then 7 (x + [x]) + 7 (x + {x} ) ≥ 3x 4 + 4 [x]4 + {x} 4 . 4
Proof. In inequality
∑ (a + b )
4
≥
4 ∑ a 4 , we take a = x, b = [x], c = {x} . 7 {x}2
Theorem 41. If x > 0 , then Proof. In inequality
2
2
2
2[x] 2x + + x x + [x]
2a
1
≥
3 . 20
b = [x], c = {x} .
5 x 2 − 4[ x ]{ x} + ≥ . (x + {x})2 + {x}2 4 x 2 x 2 + [ x ]{x} 1
(x + [ x ])2
2
2{x} ≤ 3. x + {x}
∑ a + b ≤ 3 , we take a = x,
Theorem 43. If x > 0 , then Proof. In inequality
[x]2
+
(x + {x}) + [x] (x + [x]) + {x} (b + c − a )2 ≥ 3 , we take a = x, b = [x], c = {x} . ∑ (b + c )2 + a 2 5
Theorem 42. If x > 0 , then Proof. In inequality
4
⎛
(
⎞
(∑ xy )⎜⎝ ∑ (x +1 y ) ⎟⎠ ≥ 94 , we take y = [x] , z = {x} . 2
REFERENCES: [1] [2]
Mihály Bencze, Inequalities (manuscript), 1982. Collection of “Octogon” Mathematical Magazine (1993-2006).
7
)
[Published in OCTOGON Mathematical Magazine, Vol. 14, No. 1, pp. 206211, 2006.]
8
