Several proofs and generalizations of a fractional
inequality with constraints
Fuhua Wei and Shanhe Wu
∗
Department of Mathematics and Computer Science, Longyan University, Longyan, Fujian 364012, P. R. China
E-mail: [email protected]
∗
Corresponding Author
Abstract: Ten different proofs are given for a fractional inequality with constraints. Finally, two generalized
forms are established by introducing exponent parameters and additive terms.
Keywords: fractional inequality; Cauchy-Schwarz inequality; rearrangement inequality; arithmetic-geometric
means inequality; generalization
2000 Mathematics Subject Classification: 26D15
1
Introduction
The 2nd problem given at the 36th IMO held at Toronto (Canada) in 1995 was:
Problem 1. Let a, b, c be positive real numbers with abc = 1. Prove that
1
1
1
3
+
+
≥ .
a3 (b + c) b3 (c + a) c3 (a + b)
2
In this paper, we show several different proofs and generalized forms of the inequality (1).
2
Several proofs for the inequality (1)
Proof 1. It follows from the condition abc = 1 that
1
b2 c2
1
c2 a2
1
a2 b2
=
, 3
=
, 3
=
.
+ c)
a(b + c) b (c + a)
b(c + a) c (a + b)
c(a + b)
a3 (b
Now, the inequality (1) is equivalent to
b2 c2
c2 a2
a2 b2
3
+
+
≥ .
a(b + c) b(c + a) c(a + b)
2
By Cauchy-Schwarz inequality (see [1]), we have
b2 c2
c2 a2
a2 b2
λ21 + λ22 + λ33
+
+
≥ (bc + ca + ab)2 .
λ21
λ22
λ23
Using a substitution
λ21 = a(b + c), λ22 = b(c + a), λ23 = c(a + b)
in the above inequality, and applying the arithmetic-geometric means inequality, we obtain
b2 c2
c2 a2
a2 b2
1
3p
3
+
+
≥ (bc + ca + ab) ≥ 3 (abc)2 = .
a(b + c) b(c + a) c(a + b)
2
2
2
1
(1)
Proof 2. We note that the inequality (1) is equivalent to
b2 c2
c2 a2
a2 b2
3
+
+
≥ .
ab + ac bc + ba ca + cb
2
Let
b2 c2
c2 a2
a2 b2
+
+
.
ab + ac bc + ba ca + cb
Using Cauchy-Schwarz inequality and arithmetic-geometric means inequality gives
K=
[(ab + ac) + (bc + ba) + (ca + cb)] K
≥
=
√
ab + ac · √
√
√
bc
ca
ab
+ bc + ba · √
+ ca + cb · √
ab + ac
bc + ba
ca + cb
2
(bc + ca + ab)2
p
≥ 3(bc + ca + ab) 3 (bc)(ca)(ab)
= 3(bc + ca + ab).
Hence K ≥ 32 . The desired conclusion follows.
Proof 3. Note that for a > 0,
a+
1
1
≥ 2 ⇐⇒ a ≥ 2 − .
a
a
We thus have
1
2
1
1
a2 (b + c)
1 ab + ac
=
≥
2−
= −
.
3
2
a (b + c)
2a a (b + c)
2a
2
a
4
Similarly
1
1 bc + ba
≥ −
,
b3 (c + a)
b
4
1 ca + cb
1
≥ −
.
+ b)
c
4
c3 (a
Adding the above inequalities yields
1
1
1
+
+
a3 (b + c) b3 (c + a) c3 (a + b)
1 1 1 1
+ + − (ab + bc + ca)
a b
c 2
1
= (ab + bc + ca).
2
Finally, the arithmetic-geometric means inequality leads us to the required inequality.
≥
Proof 4. The inequality (1) is equivalent to
b2 c2
c2 a2
a2 b2
3
+
+
≥ .
a(b + c) b(c + a) c(a + b)
2
On the other hand, we have for λ > 0,
2
√
b2 c2
+ λa(b + c) ≥ 2 λbc,
a(b + c)
√
c2 a2
+ λb(c + a) ≥ 2 λca,
b(c + a)
√
a2 b2
+ λc(a + b) ≥ 2 λab.
c(a + b)
Adding the above inequalities yields
√
≥ (2 λ − 2λ)(ab + bc + ca)
q
√
3
2
≥ 6( λ − λ) (abc)
√
= 6( λ − λ).
c2 a2
a2 b2
b2 c2
+
+
a(b + c) b(c + a) c(a + b)
Choosing λ =
1
4
gives
b2 c2
c2 a2
a2 b2
3
+
+
≥ ,
a(b + c) b(c + a) c(a + b)
2
which is the required inequality.
Proof 5. We make the substitution bc = x, ca = y, ab = z, x + y + z = s. Then
1
1
1
+
+
+ c) b3 (a + c) c3 (a + b)
a3 (b
=
=
x2
y2
z2
+
+
y+z
z+x x+y
2
x
y2
z2
+
+
.
s−x s−y s−z
We consider the probability distribution sequence of random variable ξ below:
s−x
y
s−y
z
s−z
x
=
, p ξ=
=
, p ξ=
=
.
p ξ=
s−x
2s
s−y
2s
s−z
2s
It follows that
x
s−x
y
s−y
z
s−z
x+y+z
1
·
+
·
+
·
=
= ,
s−x
2s
s−y
2s
s−z
2s
2s
2
2
2
x
s−x
1
x
y2
z2
Eξ 2 =
+ ( − z) =
+
+
.
s−x
2s
2s s − x s − y s − z
Eξ =
2
According to D(ξ) = Eξ 2 − (Eξ) > 0, we have
2
1
x
y2
z2
1
+
+
≥ ,
2s s − x s − y s − z
4
so
x2
y2
z2
1
1
3√
3
+
+
≥ s = (x + y + z) ≥ 3 xyz = .
s−x s−y s−z
2
2
2
2
Hence
1
1
1
3
+
+
≥ .
a3 (b + c) b3 (a + c) c3 (a + b)
2
3
Proof 6. Let bc = x, ca = y, ab = z. The inequality (1) is equivalent to
y2
z2
3
x2
+
+
≥ .
y+z
z+x x+y
2
By symmetry, we may assume that x ≥ y ≥ z, then
x
y
z
≥
≥
.
y+z
z+x
x+y
Using the rearrangement inequality (see [2]) gives
y2
z2
x
y
z
x2
+
+
≥z·
+x·
+y·
,
y+z
z+x x+y
y+z
x+z
x+y
x2
y2
z2
x
y
z
+
+
≥y·
+z·
+x·
,
y+z
z+x x+y
y+z
z+x
x+y
Adding the above inequalities yields
2
y2
z2
x
√
+
+
≥ x + y + z ≥ 3 3 xyz = 3,
2
y+z
z+x x+y
The required inequality follows.
Proof 7. Apply the same substitution as in Proof 6. The inequality (1) is equivalent to
y2
z2
3
x2
+
+
≥ .
y+z
z+x x+y
2
1
Since (x2 , y 2 , z 2 ) and ( y+z
,
inequality that
1
1
z+x , x+y )
are similarly sorted sequences, it follows from the rearrangement
x2
y2
z2
1
+
+
≥
y+z
z+x x+y
2
y2 + z2
z 2 + x2
x2 + y 2
+
+
y+z
z+x
x+y
.
By the power mean inequality, we have
y z z x x y 16
z 2 + x2
x2 + y 2
y+z
z+x x+y
y2 + z2
+
+
≥
+
+
≥6
· · · · ·
= 3,
y+z
z+x
x+y
2
2
2
2 2 2 2 2 2
this yields
x2
y2
z2
3
+
+
≥ .
y+z
z+x x+y
2
Proof 8. Let
1
a
+
1
b
+
1
c
= t (t > 0), then
1
1
1
+
+
a3 (b + c) b3 (a + c) c3 (a + b)
=
a−2
b−2
c−2
+
+
b−1 + c−1
c−1 + a−1
a−1 + b−1
=
a−2
b−2
c−2
+
+
.
−1
−1
t−a
t−b
t − c−1
Consider the following function:
g(x) =
x2
t−x
(0 < x < t).
4
Since
g 00 (x) =
2t2
3
(t − x)
>0
(0 < x < t),
we conclude that the function g is convex on (0, t).
Using Jensen’s inequality gives
a−1 + b−1 + c−1
)
g(a−1 ) + g(b−1 ) + g(c−1 ) ≥ 3g(
3 1 1 1 1
=
+ +
2 a b
c
r
33 1 1 1
≥
· ·
2 a b c
3
=
.
2
Proof 9. Consider the following function:
2 2 2
√
√
√
ca
ab
bc
x − ab + ac + √
x − bc + ab + √
x − ac + bc
f (x) = √
ab + ac
bc + ab
ac + bc
2 2
b c
c2 a2
a2 b2
=
+
+
x2 − 2 (ab + bc + ca) x + 2 (ab + bc + ca) .
ab + ac bc + ab ac + bc
Since f (x) ≥ 0 for x ∈ R, we have the discriminant ∆ ≤ 0, that is
2 2
b c
c2 a2
a2 b2
2
4 (ab + bc + ac) − 8
+
+
(ab + bc + ac) ≤ 0.
ab + ac bc + ab ac + bc
Thus
b2 c2
c2 a2
a2 b2
1
3√
3
3
+
+
≥ (ab + bc + ac) ≥
a2 b2 c2 = ,
ab + ac bc + ab ac + bc
2
2
2
which leads to
1
1
1
3
+
+
≥ .
a3 (b + c) b3 (c + a) c3 (a + b)
2
Proof 10. Construct the following vectors:
p
p
p
−→
OA =
a (b + c), b (c + a), c (a + b) ,
!
−−→
bc
ca
ab
p
OB =
, p
, p
.
a (b + c)
b (c + a)
c (a + b)
−→
−−→
We denote by θ (0 ≤ θ ≤ π) the angle of vectors OA and OB.
Since
−→ p
OA = 2 (ab + bc + ca),
−−→
OB =
s
b2 c2
c2 a2
a2 b2
+
+
,
a (b + c) b (c + a) c (a + b)
5
we have
−→ −−→ −→ −−→
OA · OB = OA OB cos θ
s
p
b2 c2
c2 a2
a2 b2
= 2 (ab + bc + ca)
+
+
cos θ
a (b + c) b (c + a) c (a + b)
s
p
b2 c2
c2 a2
a2 b2
≤ 2 (ab + bc + ca)
+
+
.
a (b + c) b (c + a) c (a + b)
On the other hand, we have
−→ −−→
OA · OB = ab + bc + ca.
Thus
c2 a2
a2 b2
1
3√
3
b2 c2
3
+
+
≥ (ab + bc + ca) ≥
a2 b2 c2 = ,
a (b + c) b (c + a) c (a + b)
2
2
2
which leads us to the inequality (1).
3
Generalizations of the inequality (1)
Theorem 1. Let a, b, c be positive real numbers such that abc = 1, and let λ ≥ 2. Then
1
1
1
3
+
+
≥ .
aλ (b + c) bλ (c + a) cλ (a + b)
2
Proof. Let a = x1 , b = y1 , c = z1 . Then
1
1
1
xλ−1
y λ−1
z λ−1
+ λ
+ λ
=
+
+
,
+ c) b (c + a) c (a + b)
y+z
z+x x+y
aλ (b
By symmetry, we may assume that x ≥ y ≥ z, then
xλ−2 ≥ y λ−2 ≥ z λ−2 ,
and
1
1
1
≥
≥
y+z
z+x
x+y
xλ−2
y λ−2
z λ−2
≥
≥
.
y+z
z+x
x+y
Using the rearrangement inequality gives
xλ−1
y λ−1
z λ−1
xλ−2
y λ−2
z λ−2
+
+
≥z·
+x·
+y·
,
y+z
z+x x+y
y+z
z+x
x+y
xλ−1
y λ−1
z λ−1
xλ−2
y λ−2
z λ−2
+
+
≥y·
+z·
+x·
.
y+z
z+x x+y
y+z
z+x
x+y
Adding the above inequalities yields
xλ−1
y λ−1
z λ−1
+
+
y+z
z+x x+y
≥
≥
=
6
1 λ−2
(x
+ y λ−2 + z λ−2 )
2
3p
3
xλ−2 y λ−2 z λ−2
2
3
.
2
(2)
The inequality (2) is proved.
Theorem 2. Let x1 , x2 , . . . , xn be positive real numbers such that x1 x2 · · · xn = 1, and let n ≥ 3, λ ≥ 3.
Then
1
xi )λ−1 (
X
Q
(
1≤k<l≤n
1≤i≤n, i6=k,l
P
≥
xi xj )
n(n − 1)
.
(n + 1)(n − 2)
(3)
1≤i<j≤n, i6=k,l
Proof. Applying the generalized Radon’s inequality (see [3-6]):
n
X
ap
i
i=1
bi
n
n
X
X
≥ n2−p (
ai )p /(
bi )
i=1
i=1
(ai > 0, bi > 0, i = 1, 2, . . . , n, p ≥ 2 or p ≤ 0), we deduce that
1
X
1≤k<l≤n
Q
(
1≤k<l≤n
[(
1
xi )/xk xl )]λ−1 [(
1≤k<l≤n
Q
1≤i≤n
(
P
xi xj ) − xk xl )]
1≤i<j≤n
1≤i<j≤n
n(n − 1)
≥
2
n(n − 1)
=
2
xk xl )λ−1
P
(
3−λ
1≤k<l≤n
P
1≤k<l≤n
xi xj )
(x x )λ−1
P k l
xi xj ) − xk xl
X
=
P
1≤i<j≤n, i6=k,l
1≤i≤n, i6=k,l
X
=
xi
)λ−1 (
3−λ P
[(
xi xj ) − xk xl )]
1≤i<j≤n
n(n − 1)
−1
2
−1
(
X
xk xl )λ−2
1≤k<l≤n
λ−2
3−λ
Y
2
2
n(n − 1)
n(n
−
1)

(
xk ) n 
≥ 2
n −n−2
2
2

1≤k≤n
=
n(n − 1)
.
(n + 1)(n − 2)
This completes the proof of Theorem 2.
Remark. In a special case when n = 3, λ = 3, x1 = a, x2 = b, x3 = c, the inequality (3) would reduce
to the inequality (1).
Acknowledgements. The present investigation was supported, in part, by the innovative experiment
project for university students from Fujian Province Education Department of China under Grant No.214,
and, in part, by the innovative experiment project for university students from Longyan University of China.
7
References
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[3] Sh.-H. Wu, An exponential generalization of a Radon inequality, J. Huaqiao Univ. Nat. Sci. Ed., 24 (1)
(2003), 109–112.
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Ed., 22 (1) (2004), 1–4.
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8