ISSN 1222-5657, ISBN 978-973-88255-5-0, www.hetfalu.ro/octogon
On Bergstr¨ om’s inequality involving six numbers
Jian Liu22
ABSTRACT.In this paper, we discuss the Bergst¨om inequality involving six numbers, four refinements and two reverse inequalities are proved. Finally, two conjectures are put forward, one of them is actually generalization of Schur’s inequality.
1. INTRODUCTION
Letx, y, z be real numbers and letu, v, w be positive numbers, then x2
u +y2 v +z2
w ≥ (x+y+z)2
u+v+w , (1.1)
with equality if and only ifx:y:z=u:v:w.
Inequality (1.1) is a special case of the Bergstr¨om inequality (see [1]-[5]).
Also it is a corollary of Cauchy-Buniakowsky-Schwarz inequality. In this paper, we will prove its four refinements and two reverse inequalities.
In the proofs of the following theorems, we denote cyclic sum on x, y, z, u, v, w by P
, for instance, Xu=u+v+w,Xx2
u = x2 u +y2
v +z2 w,X
(vz−wy)2=
= (vz−wy)2+ (wx−uz)2+ (uy−vx)2. 2. FOUR REFINEMENTS
Firstly, we give the following refinement of Bergstr¨om inequality (1.1):
22Received: 21.02.2009
2000Mathematics Subject Classification. 26D15
Key words and phrases. Bergstr¨om‘s inequality; real numbers; positive numbers;
Schur‘s inequality.
Theorem 1. Letx, y, z be real numbers and let u, v, w be positive numbers.
Then x2
u +y2 v +z2
w ≥ (x+y+z)2
u+v+w +(vz−wy)2
2vw(v+w)+(wx−uz)2
2wu(w+u)+(uy−vx)2 2uv(u+v), (2.1) with equality if and only ifx:y:z=u:v:w.
Proof. It is easy to check that X (yw−zv)
vw(v+w) = 2Xx2
u −X(y+z)2
v+w , (2.2)
Using the identity, we see that inequality (2) is equivalent to 1
2
X(y+z)2 v+w ≥ (P
x)2 Pu ,
which follows from (1.1) by replacingx→y+z, u→v+w etc., Thus inequality (2) is proved.
Clearly, equality in (2.1) holds if and only if
(y+z) : (z+x) : (x+y) = (v+w) : (w+u) : (u+v),namely x:y:z=u:v:wand this completes the proof of Theorem 1.
Applying inequality (1.1) to (2.1) we get x2
u +y2 v +z2
w ≥ (x+y+z)2
u+v+w + [(v−w)x+ (w−u)y+ (u−v)z]2
2[vw(v+w) +wu(w+u) +uv(u+v)]. (2.3) Further, we find the following stronger result:
Theorem 2. Letx, y, z be real numbers and let u, v, w be positive numbers.
Then x2
u +y2 v + z2
w ≥ (x+y+z)2
u+v+w + [(v−w)x+ (w−u)y+ (u−v)z]2
vw(v+w) +wu(w+u) +uv(u+v), (2.4) with equality if and only ifx:y:z=u:v:w.
Proof. We set
F1 =Xx2 u −(P
x)2 Pu −
P(vz−wy)2 Pvw(v+w). It is easy to get the following identity after some computations:
F1= a1x2+b1x+c1
uvwP uP
vw(v+w), (2.5)
where
a1=v2w2(4u2+v2+w2+ 2vw+ 2wu+ 2uv), b1 =−2uvw
wy(2u2+ 2v2−w2+vw+wu+uv) +vz(2w2+ 2u2−v2 +vw+wu+uv)],
c1 =u2
(4v2+w2+u2+ 2vw+ 2wu+ 2uv)w2y2−2vwyz(2v2+ 2w2−u2 +vw+wu+uv) + (4w2+u2+v2+ 2vw+ 2wu+ 2uv)v2z2
. Now, we shall show first that c1 >0. Note that
∆1(y)≡
−2vwz(2v2+ 2w2−u2+vw+wu+uv)2
−4
(4v2+w2+u2 +2vw+ 2wu+ 2uv)w2 (4w2+u2+v2+ 2vw+ 2wu+ 2uv)v2z2
=−24v2w2z2X uX
vw(u+v)<0, hence the following inequality:
(4v2+w2+u2+ 2vw+ 2wu+ 2uv)w2y2−
−2vwyz(2v2+ 2w2−u2+vw+wu+uv) +(4w2+u2+v2+ 2vw+ 2wu+ 2uv)v2z2>0
holds for arbitrary real numbersy, z and positive numbers u, v, w. Therefor c1 >0 is true. Since a1 >0, c1 >0 and
∆1(x)≡b21−4a1c1=−24u2v2w2(vz−wy)2X uX
vw(v+w)≤0.
We conclude that
a1x2+b1x+c1 ≥0 (2.6) holds for arbitrary real numbersx. ThusF1≥0 and (2.4) are proved.
Equality in (2.6) occurs if and only if b21−4a1c1= 0, hence the equality in (2.4) occurs iffyw−zv= 0. Because of the symmetry, we know that equality in (2.4) also occurs if and only ifzu−xw= 0, xv−uy= 0. Combining the
above argumentations, the case of equality in (2.4) holds iff x:y:z=u:v:w. This completes the proof of Theorem 2.
From the identity:
Xu3
−4X
vw(v+w) =X
u(u−v)(u−w) + 3uvw (2.7) and the special case of Schur’s inequality (see [6]):
Xu(u−v)(u−w)≥0, (2.8)
we see that
Xvw(v+w)< 1 4
Xu3
. (2.9)
Therefore, we obtain the following conclusion from Theorem 2:
Corollary 2.1. For any real numbersx, y, z and positive numbersu, v, w, we have
x2 u +y2
v +z2
w ≥ (x+y+z)2
u+v+w +4[(v−w)x+ (w−u)y+ (u−v)z]2
(u+v+w)3 . (2.10) with equality if and only ifx:y:z=u:v:w.
Remark 2.1The constant coefficient 4 in (2.10) is the best possible(We omit the proof).
In addition, it is easy to prove that
2(u3+v3+w3)≥vw(v+w) +wu(w+u) +uv(u+v) (2.11) from this and inequality (2.4) we get again
x2 u +y2
v +z2
w ≥ (x+y+z)2
u+v+w +[(v−w)x+ (w−u)y+ (u−v)z]2
2(u3+v3+w3) . (2.12) We further find this inequality can be improved as following:
Theorem 3. Letx, y, z be real numbers and let u, v, w be positive numbers.
Then
x2 u +y2
v +z2
w ≥ (x+y+z)2
u+v+w +[(v−w)x+ (w−u)y+ (u−v)z]2
u3+v3+w3 , (2.13) with equality if and only ifx:y:z=u:v:w.
Proof. Letting F2= x2
u +y2 v +z2
w −(x+y+z)2
u+v+w −[(v−w)x+ (w−u)y+ (u−v)z]2 u3+v3+w3 . By some calculations we get
F2 = a2x2+b2x+c2 uvwP
uP
u3 , (2.14)
where a2=vw
v[(w+u)w2+ (2u2+uv+v2)w+ (u+v)(u−v)2] +w(u+w)(w−u)2 , b2 =
−2uvw
[(u2+uv+v2)w+(u+v)(u−v)2]y+[v(w2+wu+u2)+(w+u)(w−u)2]z , c2 =u wm1y2+m2yz+vm3z2
, moreover, the values of m1, m2, m3 are
m1 =w
(u+v)u2+ (2v2+vw+w2)u+ (v+w)(v−w)2
+u(v+u)(u−v)2, m2 =−2vw
(v2+wv+w2)u+ (v+w)(v−w)2 , m3 =u
(v+w)v2+ (2w2+wu+u2)v+ (w+u)(w−u)2
+v(v+w)(v−w)2. First we show thatc2 >0. Since wm1 >0, vm3>0 and
∆2(y)≡(m2z)2−4(wm1)(vm3z2)
=−4uvwX uX
u3hX
u(u−v)(u−w) + 3uvwi
z2 <0, where we have used Schur’s inequality (2.8), so thatc2 >0. Taking into accounta2 >0, c2>0 and
∆2(x)≡b22−4a2c2 =−4uvw(vz−wy)2X uX
u3hX
u(u−v)(u−w) + 3uvwi
≤0, hencea2x2+b2x+c2 ≥0 is true for all real numbersx, inequality F2 ≥0 is
proved. Then, the same argument in the proof of Theorem 2 shows that,
equality in (2.13) holds if and only ifx:y:z=u:v:w. The proof of the Theorem 3 is complete.
Before giving the next result which is similar to Theorem 3, we first prove two inequalities.
Lemma 2.1. Let u, v, w be positive real numbers. Then
Xu(u2−v2)(u2−w2)≥0, (2.15) with equality if and only ifu=v =w.
Proof. For any positive real numbers u, v, w and real numberkthe following inequality holds:
Xuk(u−v)(u−w)≥0, (2.16) this is famous Schur’s inequality (see [6]). Puttingk= 12 and replacing u→u2, v →v2, w→w2, the claimed inequality follows.
Lemma 2.2. Let u, v, w be positive real numbers. Then
X(u−v)(u−w)(v+w)u2 ≥0. (2.17) with equality if and only ifu=v =w.
Proof We will use the method of difference substitution (see [7], [8]) to prove desired inequality. Without loss of generality suppose that u≥v≥w, puttingv=w+p, u=v+q(p≥0, q≥0), then plugging
v=w+p, u=w+p+q into the right-hand side of (2.17). One may easily check the identity:
X(u−v)(u−w)(v+w)u2= 2(p2+pq+q2)w3+ (2p+q)(p2+pq+ 4q2)w2+
+2q2(2p+q)2w+pq2(p+q)(2p+q), (2.18) sincep≥0, q≥0, w >0, we get the the inequality (2.17).
Now, we prove the following Theorem:
Theorem 4. Letx, y, z be real numbers and let u, v, w be positive numbers.
Then
x2 u +y2
v +z2 w ≥
≥ (x+y+z)2 u+v+w +4
(vz−wy)2+ (wx−uz)2+ (uy−vx)2
(u+v+w)3 , (2.19)
with equality if and only ifx:y:z=u:v:w.
Proof. Letting
F3≡Xx2 u −(P
x)2
Pu − 4P
(vz−wy)2 (P
u)3 .
After some computations we get the following identity F3 = a3x2+b3x+c3
uvw(u+v+w)3, (2.20) where
a3 =vw[(3v+ 3w+ 4u)vw+v(u−v)2+w(w−u)2],
b3 =−2uvw[(y+z)u2−2(v−w)(y−z)u+ (y+z)(v+w)2], c3 =u
w[(3w+ 3u+ 4v)wu+w(v−w)2+u(u−v)2]y2
−2vwyz[u2+ 2(v+w)u+ (v−w)2]
+v[(3u+ 3v+ 4w)uv+u(w−u)2+v(v−w)2]z2}. To proveF3≥0, we first prove thatc3 >0, it suffices to show that
w
(3w+ 3u+ 4v)wu+w(v−w)2+u(u−v)2 y2−
−2vwyz
u2+ 2(v+w)u+ (v−w)2 + +v
(3u+ 3v+ 4w)uv+u(w−u)2+v(v−w)2
z2 >0. (2.21) Note that
∆3(y)≡
−2vwz[u2+ 2(v+w)u+ (v−w)2] 2
−4vw
(3w+ 3u+ 4v)wu+w(v−w)2+u(u−v)2
[(3u+ 3v+ 4w)uv +u(w−u)2+v(v−w)2
=−4uvwM z2,
where M =X
u5+X
(v+w)u4−2X
(v+w)v2w2+ 20uvwX
u2+ 6uvwX vw.
Again, it is easy to verify the the following identity:
Xu5+X
(v+w)u4−2X
(v+w)v2w2+ 3uvwX vw=
=X
u(u2−v2)(u2−w2) +X
(u−v)(u−w)(v+w)u2+ 2uvwX
u2. (2.22) By Lemma 2.1 and Lemma 2.2 we see thatM >0, hence ∆3(y)<0, thus inequality (2.21) holds for all real numbers y, z and positive numbers u, v, w.
Since a3 >0, c3>0 and
∆3(x)≡b23−4a3c3=−4uvwM(vz−wy)2 ≤0.
Thusa3x2+b3x+c3 ≥0 holds for arbitrary real numbersx. So, the inequality of Theorem 4 is proved. As in the proof of the Theorem 2, we conclude that equality (2.19) holds if and only if x:y:z=u:v:w. Our proof is complete.
Remark 2.2Applying the method of undetermined coefficients, we can easily prove that the constant coefficient 4 in the right hand side of the inequality of Theorem 4 is the best possible. In addition, the analogous constant coefficients of others five Theorems in this paper are all the best possible.
TWO REVERSE INEQUALITIES
In this section, we will establish two inverse inequalities of Bergstr¨om’s Inequality (1.1).
Theorem 5. Letx, y, z be real numbers and let u, v, w be positive numbers.
Then x2
u +y2 v +z2
w ≤ (x+y+z)2
u+v+w +(vz−wy)2
vw(v+w) +(wx−uz)2
wu(w+u) +(uy−vx)2 uv(u+v), (3.1) with equality if and only ifx:y:z=u:v:w.
Proof. Setting
F4= (P x)2
Pu +X(vz−wy)2
vw(v+w) −Xx2 u ,
then we get the following identity:
F4 = a4x2+b4x+c4
uvw(v+w)(w+u)(u+v), (3.2) where
a4 = (v+w)(v+w+ 2u)v2w2,
b4 =−2uvw(v+w)(yw2+zv2+uyw+uzv), c4 =u2
(w+u)(w+u+ 2v)w2y2−2vwyz(u+v)(w+u)+
+(u+v)(u+v+ 2w)v2z2 . First, we will prove thatc1 >0. Since
∆4(y)≡[−2vwz(u+v)(w+u)]2−4
(w+u)(w+u+ 2v)w2
·
·
(u+v)(u+v+ 2w)v2z2
=
=−8(v+w)(w+u)(u+v)(u+v+w)v2w2z2<0, it follows that
(u+v)(u+v+ 2w)v2z2−2vwy(u+v)(w+u)z+
+(w+u)(w+u+ 2v)y2w2 >0. (3.3) Hence c4>0. Note that againa4 >0 and
∆4(x)≡b24−4a4c4=−(u+v+w)(v+w)(w+u)(u+v)(uvw)2(yw−vz)2 ≤0, so we have a4x2+b4x+c4≥0. ThereforeF4 ≥0 and (3.1) are proved.
Equality in (3.1) holds when x:y:z=u:v:w and the proof is complete.
Remark 3.1From identity (2.2), the inequality of Theorem 5 is equivalent to
x2 u +y2
v +z2
w +(x+y+z)2
u+v+w ≥ (y+z)2
v+w +(z+x)2
w+u +(x+y)2
u+v (3.14) Remark 3.2Combining Theorem 1 and Theorem 5, we get the following double inequalities:
(vz−wy)2
2yz(y+z) + (wx−uz)2
2zx(z+x) + (uy−vx)2 2xy(x+y) ≤ x2
u +y2 v +z2
w −(x+y+z)2 u+v+w ≤
≤ (vz−wy)2
yz(y+z) +(wx−uz)2
zx(z+x) +(uy−vx)2
xy(x+y). (3.5)
In the sequel we give another reverse inequality:
Theorem 6. Letx, y, z be real numbers and let u, v, w be positive numbers, then
x2 u +y2
v +z2
w ≤ (x+y+z)2
u+v+w +(vz−wy)2+ (wx−uz)2+ (uy−vx)2
2uvw , (3.6)
with equality if and only ifx:y:z=u:v:w.
Proof. Letting
F5 = (P x)2 Pu −
P(vz−wy)2
2uvw −Xx2 u ,
then we have the following identity:
F5= a5x2+b5x+c5
2uvw(u+v+w), (3.7)
where
a5 =u(v2+w2) + (v+w)(v−w)2, b5=−2u[u(zw+vy) + (v−w)(vy−zw)], c5=
v(w2+u2) + (w+u)(w−u)2
y2−2vwz(v+w−u)y +
w(u2+v2) + (u+v)(u−v)2 z2.
First we prove that c5>0. Since
∆5(y)≡[−2vwz(v+w−u)]2−4
v(w2+u2) + (w+u)(w−u)2 w(u2+v2) +(u+v)(u−v)2
z2
=−4u2z2hX
u3−X
(v+w)u2+ 4uvwi X u
=−4u2z2hX
u(u−v)(u−w) +uvwi X u <0,
where we have applied Schur’s inequality (2.8). Therefore, we know that
∆5(y)<0 holds for arbitrary real numbersy, z. Hencec5 >0 is true. On the other hand, sincea5 >0, c5 >0 and
∆5(x)≡b25−4a5c5=−4(vz−wy)2hX
u(u−v)(u−w) +uvwi X u≤0, thusa5x2+b5x+c5≥0 holds for arbitrary real numberx. InequalityF5≥0 and (3.6) are proved. Clearly, the equality in (3.6) holds if and only if
x:y:z=u:v:w. Hence Theorem 6 has been proved completely.
LetABC be an acute triangle. In (3.6) we take
u= tanA, v= tanB, w= tanC, multiplying by tanAtanBtanC in both sides, then using the well known identity:
tanA+ tanB+ tanC= tanAtanBtanC, (3.8) we get
Corollary 3.1. Let ABC be an acute triangle and x, y, z >0, then x2tanBtanC+y2tanCtanA+z2tanAtanB ≤(x+y+z)2+
+1
2[(ytanC−ztanB)2+ (ztanA−xtanC)2+ (xtanB−ytanA)2], (3.9) with equality if and only ifx:y:z= tanA: tanB : tanC.
4. TWO CONJECTURES
Finally, we propose two conjectures. The first is about the triangle:
Conjecture 4.1. Letx, y, z be positive real numbers and leta, b, c be the sides of △ABC, then
(b+c)2
y+z +(c+a)2
z+x +(a+b)2
x+y ≥ 2(a+b+c)2
x+y+z +(bz−cy)2
4(y+z)3 + (cx−az)2 4(z+x)3 +
+(ay−bx)2
4(x+y)3 (4.1)
with equality if and only ifx:y:z=a:b:c.
The second is a generalization of Lemma 2.2.
Conjecture 4.2. Letk be arbitrary real numbers and letx, y, z, p, q be real numbers such thatx >0, y >0, z >0, p≥0, p≥q+ 1. Then
X(xk−yk)(xk−zk)(yq+zq)xp≥0, (4.2) X(xk−yk)(xk−zk)(y+z)qxp ≥0. (4.3) If we take k= 1, q= 1, p= 2 in the conjecture, then both (4.2) and (4.3) become the inequality (2.18) of Lemma 2.2. If we putq = 0, then we obtain actually Schur’s inequality (2.16) from (4.2) or (4.3).
REFERENCES
[1] Bergstr¨om, H.,Triangle inequality for matrices, Den Elfte Skandinaviske Matematikerkongress, Trodheim, 1949, Johan Grundt Tanums Forlag, Oslo, 1952, 264-267.
[2] Bellman, R.,Notes on Matrix Theory-IV(An Inequality Due to Bergstr¨om), Amer.Math.Monthly, Vol.62(1955)172-173.
[3] Fan, Ky,Generalization of Bergstr¨om’s inequality, Amer.Math.Monthly, Vol.66,No.2(1959),153-154.
[4] Beckencbach, E.F., and Bellman, R., Inequalities, Springer, Berlin,G¨ottingen and Heidelberg, 1961.
[5] M˘arghidanu, D., D´ıaz-Barrero, J.L., and R˘adulescu, S.,New refinements of some classical inequalities, Mathematical Inequalities Applications, Vol.12, 3(2009), 513-518.
[6] Hardy, G.H., Littlewood, J.E., and P´olya, G.,Inequalities, Cambridge, 1934.
[7] Yang, L., Difference substitution and automated inequality proving, J.Guangzhou Univ. (Natural Sciences Edition), 5(2)(2006), 1-7. (in Chinese)
[8] Yu-Dong Wu, Zhi-hua Zhang, and Yu-Rui Zhang,Proving inequalities in acute triangle with difference substitution, Inequal.Pure Appl. Math., 8(3)(2007), Art.81.
East China Jiaotong University, Nanchang City, Jiangxi
Province,330013,P.R.China
