About an Identity and its Applications
Mihály Bencze and Florentin Smarandache
Str. Hărmanului 6, 505600 Săcele, Jud. Braşov, Romania
University of New Mexico, 200 College Road, Gallup, NM 87301, USA
3
3
4
Theorem 1. If x, y ∈ C then 2 ( x 2 + y 2 ) − ( x + y ) ( x 3 + y 3 ) = ( x − y ) ( x 2 + xy + y 2 ) .
Proof. With elementary calculus.
Application 1.1. If x, y ∈ C then
( 2 ( x + y ) − ( x + y ) ( x + y )) ( 2 ( x + y ) − ( x − y ) ( x − y )) = ( x − y ) ( x + x y + y )
2
2 3
3
3
3
2 3
2
3
3
3
2 4
2
4
2
2
4
Proof. In Theorem 1 we replace y → − y , etc.
Application 1.2. If x ∈ R then
4
( sin x − cos x ) (1 + sin x cos x ) + ( sin x + cos x )
3
( sin
3
x + cos3 x ) = 2
Proof. In Theorem 1 we replace x → sin x, y → cos x
3
(
4
)
(
)
Application 1.3. If x ∈ R then 2ch6 x − (1 + shx ) 1 + sh3 x = (1 + shx ) shx + ch 2 x .
Proof. In Theorem 1 we replace x → 1, y → shx
Application 1.4. If x, y ∈ C ( x ≠ ± y ) then
3
3
3
2 ( x 2 + y 2 ) − ( x + y ) ( x3 + y 3 )
( x − y)
4
+
3
2 ( x 2 + y 2 ) − ( x − y ) ( x3 − y 3 )
( x + y)
4
= 2 ( x2 + y 2 )
Application 1.5. If x, y ∈ C then
3
3
2
2
2 ( x 2 + y 2 ) − ( x + y ) ( x3 + y 3 )
x + xy + y
+
3
3
2
2
2 ( x 2 + y 2 ) − ( x − y ) ( x3 − y 3 )
x − xy + y
3
= 2 ( x4 + 6 x2 y 2 + y 4 )
3
Application 1.6. If x, y ∈ R then 2 ( x 2 + y 2 ) ≥ ( x + y ) ( x 3 + y 3 ) .
(See Jószef Sándor, Problem L.667, Matlap, Kolozsvar, 9/2001.)
Proof. See Theorem 1.
3
3
Theorem 2. If x, y , z ∈ R then 3 ( x 2 + y 2 + z 2 ) ≥ ( x + y + z ) ( x 3 + y 3 + z 3 ) .
Proof. With elementary calculus.
Application 2.1. Let ABCDA1 B1C1 D1 be a rectangle parallelepiped with sides a, b, c and
3
(
)
diagonal d . Prove that 3d 6 ≥ ( a + b + c ) a3 + b3 + c3 .
Application 2.2. In any triangle ABC the followings hold:
3
1) 3 ( p 2 − r 2 − 4 Rr ) ≥ 2 p 4 ( p 2 − 3r 2 − 6 Rr )
3
2) 3 ( p 2 − 2r 2 − 8 Rr ) ≥ p 4 ( p 2 − 12 Rr )
(
2
3) 3 ( 4 R + r ) − 2 p 2
)
3
≥ ( 4R + r )
3
(( 4R + r ) − 12 p R )
3
2
3
4) 3 ( 8 R 2 + r 2 − p 2 ) ≥ ( 2 R − r )
(
2
5) 3 ( 4 R + r ) − p 2
)
3
3
≥ ( 4R + r )
(( 2R − r ) (( 4R + r ) − 3 p ) + 6Rr )
2
3
2
2
(( 4R + r ) − 3 p ( 2R + r ))
3
2
Proof. In Theorem 2 we take:
{ x, y , z } ∈
⎧
A
B
C⎫ ⎧
A
B
C ⎫⎫
⎧
∈ ⎨{a, b, c} ; { p − a, p − b, p − c} ; {ra , rb , rc } ; ⎨sin 2 ,sin 2 ,sin 2 ⎬ ; ⎨cos 2 , cos 2 , cos 2 ⎬⎬
2
2
2⎭ ⎩
2
2
2 ⎭⎭ .
⎩
⎩
Application 2.3. Let ABC be a rectangle triangle, with sides a > b > c then
3
24a 6 ≥ ( a + b + c ) ( a3 + b3 + c3 )
⎛ n
⎞ ⎛ n
⎞
Theorem 3. If xk > 0, k = 1, 2,..., n , then n ⎜ ∑ xk2 ⎟ ≥ ⎜ ∑ xk ⎟
⎝ k =1 ⎠ ⎝ k =1 ⎠
3
∑ (C )
n
Application 3.1 The following inequality is true:
3
k
n
k =0
3
∑x
n
3
k
.
⎛ C2nn ⎞
≤ ( n + 1) ⎜
⎟ .
⎝ 2 ⎠
k =1
3
Proof. In Theorem 3 we take xk = Cnk , k = 0,1, 2,..., n. .
Application 3.2. In all tetrahedron ABCD holds:
⎛
1 ⎞
⎜∑ 2 ⎟
ha ⎠
4
1) ⎝
≥ 3
1
r
∑ h3
a
⎛
1⎞
⎜∑ 2 ⎟
ra ⎠
2
2) ⎝
≥ 3
1
r
∑ r3
a
1
1
1
1
Proof. In Theorem 3 we take x1 = , x2 = , x3 = , x4 =
and
ha
hb
hc
hd
1
1
1
1
x1 = , x2 = , x3 = , x4 = .
ra
rb
rc
rd
3
3
Application 3.3. If S nα = ∑ k α then n ( S n2α ) ≥ ( S nα ) S n3α .
n
3
3
k =1
Proof. In Theorem 3 we take xk = k α , k = 0,1, 2,..., n. .
⎛ FF ⎞
Application 3.4. If Fk denote Fibonacci numbers, then ∑ F ≤ n ⎜ n n +1 ⎟ .
k =1
⎝ Fn + 2 − 1 ⎠
Proof. In Theorem 3 we take xk = Fk , k = 1, 2,..., n. .
3
n
3
k
References:
[1]
Mihály Bencze, Inequalities (manuscript), 1982.
[2]
Collection of “Octogon Mathematical Magazine”, 1993-2004.
{Published in “Octogon Mathematical Journal”, Vol. 13, No. 1, 2005.}
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