aa r X i v : . [ m a t h . C A ] S e p The Damascus Inequality
Fozi M. Dannan, Sergey M. Sitnik
In 2016 Prof. Fozi M. Dannan from Damascus, Syria proposed the nextinequality x − x − x + 1 + y − y − y + 1 + z − z − z + 1 ≤ , (1)providing that xyz = 1 for x, y, z > . It became widely known but was notproved yet in spite of elementary formulation.An obvious generalization is the next inequality n X k =1 x k − x k − x k + 1 ≤ , (2)providing that x · x . . . · x n = 1 for x k ≥ , ≤ k ≤ n .It is obvious that (2) is true for n = 1, it is easy to prove it also for n = 2directly. But it is not true for n = 4 as follows from an example with x = x = x = 2 , x = , then (2) is reducing to 1 − ≤ n ≥ x = x = x = 2 , x = , x = . . . = x n = 1. So the only non–trivial case in(2) is n = 3.In this paper we prove inequality (1) together with similiar ones1 x − x + 1 + 1 y − y + 1 + 1 z − z + 1 ≤ xx − x + 1 + yy − y + 1 + zz − z + 1 ≤ x − x + x + 1 + y − y + y + 1 + z − z + z + 1 ≤ x + x + 1 + 1 y + y + 1 + 1 z + z + 1 ≥ xx + x + 1 + yy + y + 1 + zz + z + 1 ≤ x + 1 x + x + 1 + y + 1 y + y + 1 + z + 1 z + z + 1 ≤ . (8)Also some generalizations will be considered.1 Proof of the main inequality (1)
Theorem 1.
An inequality (1) holds true providing that xyz = 1 for x, y, z > . For the proof we need an auxiliary inequality that seems to be very inter-esting by itself.
Lemma 1.
Let x, y, z be positive numbers such that xyz = 1. Then x + y + z − x + y + z ) + 6 ≥ xyz = 1. Due to its impor-tance we give three proofs to it based on different ideas. First proof of Lemma 1.
To prove (9) let introduce the Lagrange function L ( x, y, z, λ ) = x + y + z − x + y + z ) + 6 − λ ( xyz − . On differentiating it follows λ = x − x = y − y = z − z. It follows that at the minimum (it obviously exists) x = y , so three variablesat the minimum are x, y = x, z = 1 /x . From x − x = z − z we derive theequation in x :2 x − x = 2 /x − /x , f ( x ) = 2 x − x + 3 x − . One root is obvious x = 1. Let us prove that there are no other roots for x ≥ f ′ ( x ) = 12 x − x + 6 x = 3 x (4 x − x + 2) ≥ , x ≥ . Define a function g ( x ) = 4 x − x + 2, its derivative g ′ ( x ) has one zero for x ≥ x = 15 /
16 and the function g ( x ) is positive at this zero at its minimum g (15 /
16) = 15893 / >
0. So g ( x ) is positive, f ( x ) is strictly increasing on x ≥
0, so f (1) = 0 is its only zero. Second proof of Lemma 1.
Consider the function f ( x, y ) = x + y + 1 x y − (cid:18) x + y + 1 xy (cid:19) + 6 , where x, y, z are positive numbers. We show that f ( x, y ) attains its minimum0 at x = 1 , y = 1 using partial derivative test.Calculate ∂f∂x = 2 x − x y − x y = 0 , (10) ∂f∂y = 2 y − x y − xy = 0 . (11)2ow multiplying (10) and (11) respectively by x and − y and adding to obtain( x − y ) (2 x + 2 y −
3) = 0 . Here we have two cases.Case I. x = y , which implies from equation (10) that2 x − − x + 3 x = 0or (cid:0) x − (cid:1) (cid:0) x − x + 2 (cid:1) = 0 . (12)Equation (12) has only one positive root x = 1 and consequently y = 1 . Noticethat the equation 2 x − x + 2 = 0does not have positive roots because for x ≥ u ( x ) = 2 x − x + 2satisfies the following properties : ( i ) u (0) = 2 , ( ii ) min u ( x ) = 1 at x = 1,( iii ) u ( ∞ ) = ∞ . Therefore f ( x, y ) attains its maximum or minimum at x =1 , y = 1 . Case II. 2 x + 2 y = 3 . Adding (10) and (11) we get2 ( x + y ) − −
2( 1 x y + 1 x y ) + 3 (cid:18) x y + 1 xy (cid:19) = 0or − − x y + 32 x y = 0and − x y + 3 xy − . Putting t = xy we obtain 2 t − t + 2 = 0 . In fact this equation does not have positive root (notice that t = xy should bepositive). This is because the function u = 2 t − t + 2 satisfies the followingproperties : ( i ) u (0) = 2 , ( ii ) for t > , min u ( t ) = u (cid:18) √ (cid:19) > , ( iii ) u ( ∞ ) > . The last step is to show that f ( x, y ) ≥ f (1 ,
1) = 0 . It is enough to show that f ( s, t ) > f (1 ,
1) for at least one point ( s, t ) = (1 ,
1) .Take for example f (2 ,
3) = + . Third proof of lemma 1 (Geometrical Method).
Geometrically it is enough to prove that the surface xyz = 1 lies outside thesphere ( x − / + ( y − / + ( z − / = 3 / Let M and S be surfaces defined by M : xyz = 1 and S : ( x − ) + ( y − ) + ( z − ) − = 01. If ( z − ) − ≥ x −
32 ) + ( y −
32 ) + ( z −
32 ) − ≥ x + y + z − x + y + z ) + 6 ≥ .
2. If ( z − ) − ≤ − √ ≤ z ≤ √ .
3. We take horizontal sections for both M and so get for any plane3 − √ ≤ z = k ≤ √ H ( k ) with vertex ( √ k , √ k ) and a circle C ( k )which radius is given by r ( k ) = 34 − ( k −
32 ) = − k + 3 k − , , k ).4. For z = 1 , we have the hyperbola xy = 1 and the circle( x −
32 ) + ( y −
32 ) = 12 .
5. We show that the distance d ( v, c ) between the vertex of the hyperbolaand the center c of the circle is always greater than or equal to the radius ofthe circle. The distance d ( v, c ) is given by d ( v, c ) = (cid:18) √ k − (cid:19) + (cid:18) √ k − (cid:19) = 2 (cid:18) √ k − (cid:19) . r ( k ) = 34 − ( k −
32 ) = − k + 3 k − . We need to show that the vertex is always outside the circle i.e. d ( v, c ) ≥ r ( k )for all 3 − √ ≤ k ≤ √ . Clearly that d ( v, c ) = r for k = 1 and the hyperbola tangents the circle at thepoint (1 , ,
1) .For − √ ≤ k < k decreases from 1 to − √ , the radius of thecircle becomes smaller . From the other side the vertex ( √ k , √ k , k ) moves awayfrom (1 , ,
1) towards a point (0 , , k ). This follows from the distance functionof the vertex Ov = √ √ k , < k ≤ k < → √ √ k < √ √ k ! .
6. For 1 < k ≤ √ , we show that d ( v, c ) = g ( k ) = 2 (cid:18) √ k − (cid:19) > r = 34 − ( k −
32 ) = h ( k ) . In fact, h ( k ) is a concave down parabola and has its maximum at k = 1 , i.e.max h ( k ) = and h ( k ) is decreasing for k > . Also, g (1) = and g ( k ) isincreasing for k > g ′ ( k ) = 4 (cid:16) − k √ k (cid:17) (cid:16) √ k − (cid:17) > k ≥ √ k < g ( k ) > h ( k ) for k > x −
32 ) + ( y −
32 ) + ( z −
32 ) − ≥ x + y + z − x + y + z ) + 6 ≥ x, y, z ) that satisfy xyz = 1. Proof of the theorem 1.
Now consider the inequality to prove (1). After simplifying with the use ofWolfram Mathematica it reduces to − x − x + 3 y − xy + 2 x y − y + 2 xy − x y + 3 z − xz + 2 x z −− yz + 3 xyz − x yz + +2 y z − xy z + x y z − z + 2 xz − x z ++2 yz − xyz + x yz − y z + xy z ≤ . Using
SymmetricReduction function of Wolfram Mathematica we derive3 − xy − xz − yz + 3 xyz − x + y + z ) + 2( x + y + z ) − xyz ( xy + xz + yz ) − − x + y + z )( xy + xz + yz ) + ( xy + xz + yz ) ≥ . xyz = 1 let further simplify6 − x + y + z ) + 2( x + y + z ) − xy + xz + yz ) −− x + y + z )( xy + xz + yz ) + ( xy + xz + yz ) ≥ . In terms of elementary symmetric functions S = x + y + z, S = xy + yz + xz it is S − S S − S + 2 S − S + 6 ≥ . (13)As S − S S + S ≥ S − S − S + 6 ≥ . (14)Expanding it again in x, y, z we derive an inequality to prove for positivevariables x + y + z − x + y + z ) + 6 ≥ . (15)But this is exactly an inequality from Lemma 1. So Theorem 1 is proved. (3) – (8) Let us start with two propositions.
Proposition 1.
For any real numbers u, v, w such that(1 + u ) (1 + v ) (1 + w ) > , the inequality 11 + u + 11 + v + 11 + w ≤ k ( ≥ k )is equivalent to kuvw + ( k −
1) ( uv + vw + wu ) + ( k −
2) ( u + v + w ) + k − ≥ ≤ . Proposition 2.
For any real numbers u , v , w such that( u −
1) ( v −
1) ( w − > u − v − w − ≤ k ( ≥ k )is equivalent to kuvw − ( k + 1) ( uv + vw + wu ) + ( k + 2) ( u + v + w ) − ( k + 3) ≥ ≤ . The validity of propositions 1 and 2 can be obtained by direct expansions.
Proof that (1) ⇔ (3) . In fact x − x − x + 1 + y − y − y + 1 + z − z − z + 1 =6 x − ( x − x + 1) x − x + 1 + y − ( y − y + 1) y − y + 1 + z − ( z − z + 1) z − z + 1 == − X cyc x x − x + 1 . Now if the right side is ≤ X cyc x x − x + 1 ≤ X cyc − (1 /x ) + (1 /x ) ≤ . Proof of 5.
We need to prove X cyc x + x + 1 ≥ u = x + x , v = y + y , w = z + z . Using Proposition 1 the requiredinequality can be written as follows : uvw − ( u + v + w ) − ≤ . Going back to x, y, z we get( x + 1) ( y + 1) ( z + 1) − (cid:0) x + y + z (cid:1) − ( x + y + z ) − ≤ . Or xy + yz + zx ≤ x + y + z which is obvious. Proof of 6.
It follows from elementary calculus that for any real number x we have xx + x + 1 ≤ Proof that (6) + (7) ⇒ (5) . Really adding together (7) with (6) multiplied by − Proof that (6) ⇒ (8) . The required inequality is equivalent to X cyc x + x + 1 − x x + x + 1 = 3 − X cyc x x + x + 1 ≤ X cyc x x + x + 1 ≥ Modifications of original inequality
In this section we consider modifications of the original inequality (1) pro-viding that xyz = 1 for x, y, z ≥ .
1. An inequality (1) is equivalent to x − x − y − y − z − z − ≤ . (16)This form leads to generalization with more powers, cf. below.2. An inequality (1) is equivalent to x x − x + 1 + y y − y + 1 + z z − z + 1 ≤ . (17)3. Let take x → x , y → y , z → z . Then we derive another equivalent formof the inequality (1) x − x x − x + 1 + y − y y − y + 1 + z − z z − z + 1 ≤ , (18)due to the functional equation f ( 1 x ) = − xf ( x ) (19)for the function f ( x ) = x − x − x + 1 . (20)So it seems possible to generalize the original inequality in terms of functionalequations too.To one more similar variant leads a change of variables x → xy, y → yz, z → xz : xy − x y − xy + 1 + yz − y z − yz + 1 + xz − x z − xz + 1 ≤ , (21)or like (18) xy − x y x y − xy + 1 + yz − y z y z − yz + 1 + xz − x z x z − xz + 1 ≤ , (22)It is also possible to consider generalizations of (1) under the most generaltransformations x → g ( x, y, z ) , y → h ( x, y, z ) , z → g ( x,y,z ) h ( x,y,z ) with positivefunctions g ( x, y, z ) , h ( x, y, z ) still preserving a condition xyz = 1.4. A number of cyclic inequalities follow from previous ones by a substitution x = ab , y = bc , z = ca , xyz = 1 .
8n this way we derive from (1), (3)–(8) the next cyclic inequalities: ab − b a − ab + b + bc − c b − bc + c + ca − a c − ca + a ≤ , (23) b a − ab + b + c b − bc + c + a c − ca + a ≤ aba − ab + b + bcb − bc + c + cac − ca + a ≤ ab − b a + ab + b + bc − c b + bc + c + ca − a c + ca + a ≤ b a + ab + b + c b + bc + c + a c + ca + a ≥ aba + ab + b + bcb + bc + c + cac + ca + a ≤ ab + b a + ab + b + bc + c b + bc + c + ca + a c + ca + a ≤ . (29)On cyclic inequalities among which Schur, Nessbit and Shapiro ones are themost well–known cf. [1]–[3].5. Some geometrical quantities connected with trigonometric functions andtriangle geometry satisfy a condition xyz = 1, cf. [4]–[6]. For example, we mayuse in standard notations for triangular geometry values: x = a p , y = bR , z = cr ; x = a + b , y = b + cp , z = a + cp + r + 2 rR ; x = Rh a , y = h b p , z = h c r ; x = 2 R sin( α ) , y = sin( β ) r , z = sin( γ ) p ; x = ( p − R − rR − r ) tan( α ) , y = tan( β )2 p , z = tan( γ ) r ; x = tan( α )tan( α ) + tan( β ) + tan( γ ) , y = tan( β )tan( α ) + tan( β ) + tan( γ ) , z = tan( γ )tan( α ) + tan( β ) + tan( γ ) ; x = tan( α/ , y = p tan( β/ , z = tan( γ/ r ; x = a p − a ) , y = bR ( p − b ) , z = r cp − c ; x = 4 R sin( α/ , y = sin( β/ , z = sin( γ/ r ; x = 4 R cos( α/ , y = cos( β/ , z = cos( γ/ p .
6. The above geometrical identities of the type x y z = 1 which we usefor applications of considered inequalities are mostly consequences of Vieta’sformulas [5]. It is interesting to use these formulas for cubic equation directly.9 heorem 2.
Let x, y, z be positive roots of the cubic equation with anyreal a, b t + at + bt − . The for these roots x, y, z all inequalities of this paper are valid.7. We can generalize inequalities (3), (6)–(8) for more general powers. Forthis aim we use Bernoulli’s inequalities [1]–[2] : for u > u α − αu + α − ≥ , ( α > α < ,u α − αu + α − ≤ , ( 0 < α < . Lemma 2.
Assume that x, y, z are positive numbers such that xyz = 1 . Then the following inequality holds true : (cid:18) x − x + 1 (cid:19) α + (cid:18) y − y + 1 (cid:19) α + (cid:18) z − z + 1 (cid:19) α ≤ < α < . Proof . Let X = x − x + 1 , Y = y − y + 1 , Z = z − z + 1 . Then we have (cid:18) x − x + 1 (cid:19) α + (cid:18) y − y + 1 (cid:19) α + (cid:18) z − z + 1 (cid:19) α ≤≤ α (cid:18) X + 1 Y + 1 Z (cid:19) + 3 (1 − α ) ≤ . Similarly we have from (7) that (cid:18) xx + x + 1 (cid:19) α + (cid:18) yy + y + 1 (cid:19) α + (cid:18) zz + z + 1 (cid:19) α ≤ − α and from (8) we have (cid:18) x + 1 x + x + 1 (cid:19) α + (cid:18) y + 1 y + y + 1 (cid:19) α + (cid:18) z + 1 z + z + 1 (cid:19) α ≤ − α . For α > α < (cid:18) x + x + 1 (cid:19) α + (cid:18) y + y + 1 (cid:19) α + (cid:18) z + z + 1 (cid:19) α ≥ − α. It is easy to show that the maximum of the function (20) is attained for x ≥ x = 2 and equals to 1/3. 10 - So the next unconditional inequality holds k = n X k =1 x k − x k − x k + 1 ≤ n x k ≥ S = k = n X k =1 x k , S = n X k,m =1 ,k = m x k · x m , . . . , S n = x x · · · x n . The generalized Damascus inequality
Prove an inequality k = n X k =1 x k − x k − x k + 1 ≤ n − C ( a , a , · · · , a n ); x k ≥ S = a , S = a , · · · , S n = a n (32)with may be some restrictions in (32) omitted.The unconditional constant for positive numbers in (31) is C = 0 and theoriginal inequality gives C = n in case n = 3 and a single restriction S = 1 inthe list (32).It seems that a problem to find the sharp constant in the inequality (31)under general conditions (32) is a difficult problem.For three numbers so more inequalities of the type (31) may be considered,e.g.1. Prove inequality (31) for positive numbers under condition S = 1 andfind the best constant for this case.2. Prove inequality (31) for positive numbers under condition S = 1 andfind the best constant for this case.Also combined conditions may be considered.3. Prove inequality (31) for positive numbers under conditions S = a, S = b and find the best constant C ( a, b ) in (31) for this case.11 Symmetricity of symmetric inequalities
There are many inequalities that are written in terms of symmetric functionsas F ( p, q ) ≤ ≥ p = S = x + y + z, q = S = xy + yz + zx, r = S = xyz = 1 . The following Lemma enlarge the amount of inequalities that one can obtain asa series of very complicated inequalities.
Lemma 3.
If the inequality F ( p, q ) ≤ ≥ i ) F ( q, p ) ≤ ≥ , and ( ii ) F (cid:0) q − p , p − q (cid:1) ≤ ≥ . Proof. (i). Assume that F ( p, q ) = F ( x + y + z, xy + yz + zx ) ≥ . Using transformations x → xy, y → yz, z → zx we obtain F ( p, q ) = F ( xy + yz + zx, xyyz + yzzx + zxxy ) == F ( xy + yz + zx, x + y + z ) = F ( q, p ) ≥ . Notice that we can also use transformations x → x , y → y , z → z . (ii). Now assume that F ( p, q ) = F ( x + y + z, xy + yz + zx ) ≥ . Using transformations x → xyz , y → yzx , z → zxy we derive xyz + yzx + zxy = x y + y z + z x == ( xy + yz + zx ) − x + y + z ) = q − p . Also it follows xyz yzx + yzx zxy + zxy xyz == yzx + zxy + xyz = x + y + z
12 ( x + y + z ) − xy + yz + zx ) = p − q . The proof is complete.At the end we propose an unsolved problem.
Problem.