aa r X i v : . [ m a t h . G M ] J a n An Elegant Inequality
Yiguang Liu ✦ Abstract —A new inequality, ( x ) p +(1 − x ) p ≤ for p ≥ and ≥ x ≥ is found and proved. The inequality looks elegant as it integrates twonumber pairs ( x and − x , p and p ) whose summation and product areone. Its right hand side, , is the strict upper bound of the left hand side.The equality cannot be categorized into any known type of inequalitiessuch as H¨older, Minkowski etc. In proving it, transcendental equationshave been met with, so some novel techniques have been built to getover the difficulty. Index Terms —Inequality; Derivative test; Transcendental equation
NTRODUCTION I NEQUALITIES are critical for studying performance ofalgorithms, models and others. For example, inequalitiesinvolving eigenvalues and singular values of products andsums of matrices are useful to analyze multi-variable sys-tems [1]; Sobolev Type Inequalities have a strong influenceon development of the theory of partial differential equa-tions, analysis, mathematical physics, differential geometry,and other fields [2]. So, many books about inequalities havebeen published. For instance, [3] consists of more than 5000inequalities and 50 methods for proving inequalities.Recently, inequalities still keep rapid developing. Corre-lation inequalities have been particularly paid attention tofor studying stochastic models for large interacting systems[4]. Bernoullis inequality is often used as the crucial step inthe proof of other inequalities, and various generalizationsof it were given in [5]. As is well known, the H¨older’sinequality has important applications in many areas ofpure and applied mathematics, and a new sharpened andgeneralized version of H¨older’s inequality has been givenin [6]. It has also been refined and extended to the caseof multiple sequences [7]. New generalizations of Acz´el’sinequality and Popoviciu’s inequality were given [8] ,which has broad applications to mathematical analysis. In[9], a generalization of the Ky Fan discrete multivariateinequality was obtained and some of its applications werealso given.Though there are so many classical inequalities, such asfundamental inequalities, combinatorial inequalities, vari-ational inequalities, determinant and matrix inequalities,sequence and series inequalities, differential inequalities,etc, we don’t meet with an algebraic inequality which con-sists of two numbers which are reciprocal, and of anothertwo numbers whose summation is . In this paper, we putforward a formally elegant inequality having this property. • Yiguang Liu is with College of Computer Science and with School ofAeronautics and Astronautics, Sichuan University, Chengdu, SichuanProvince, China, 610065. E-mail: [email protected]
ATHEMATICAL A PPARATUS
In calculus, a derivative test is used to locate the criticalpoints of a function and determine whether such pointsare a local maximum or a local minimum. The usefulnessof derivatives to find extrema is proved mathematically byFermat’s theorem of stationary points. To make this paperself-contained, some criteria borrowed from textbooks suchas [10] are listed as follows.
Lemma 1.
Let f : ( a, b ) → R be a function and supposethat x ∈ ( a, b ) is a point where f has a local extremum.If f is differentiable at x , then ˙ f ( x ) = 0 . That is, if thederivative of a function at any point is not zero, thenthere is not a local extremum at that point. Lemma 2. If x is an extremum of f , then one of the follow-ing is true: 1) x is at a or b ; 2) f is not differentiable at x ; 3) x is a stationary point of f . Lemma 3.
If the function f is twice-differentiable at station-ary point x , then: 1) if ¨ f ( x ) < , then f attains a localmaximum at x ; 2) if ¨ f ( x ) > , then f attains a localminimum at x ; 3) if ¨ f ( x ) = 0 , then whether f ( x ) is alocal extreme is inconclusive. Using derivative test, we have proved the proposed in-equality in this section. First, we prove the following theo-rem.
Theorem 1.
For p ≥ , ( ) p + ( ) p ≤ . Proof : Let f ( p ) = ( 12 ) p + ( 12 ) p , p ≥ . (1)From (1) we can see f ( p ) is always derivative, so there are ˙ f ( p ) = − ( 12 ) p log(2) + 1 p ( 12 ) p log(2) , (2)and ¨ f ( p ) = ( 12 ) p log (2) − p ( 12 ) p log(2) + 1 p ( 12 ) p log (2) . (3)At stationary points, ˙ f ( p ) = 0 , from (2) there is ( 12 ) p = 1 p ( 12 ) p ≡ C > . (4) To determine whether f is locally maximal or locally min-imal, we need to consider the twice derivative of f . Sosubstituting (4) into (3), we have ¨ f ( p ) = C (log (2) − p log(2) + 1 p log (2)) (5) = C ( 1 p − t )( 1 p − t ) where t ≡ − √ − log (2)log(2) ≈ . < , t ≡ √ − log (2)log(2) ≈ . . In terms of the properties ofquadratic functions, and p ≥ , we have ¨ f ( p ) ( ≤ when p ∈ [ t , > when p ∈ (0 , t ) . That is, when p ∈ [1 , t − ] ≈ [1 , . , there is ¨ f ( p ) ≤ ,which means f p take local maximal values at the stationarypoints ˙ f ( p ) = 0 ; when p > t − , ¨ f ( p ) > means f p takelocal minimal values at ˙ f ( p ) = 0 . Because we only considerthe maximal value of function f ( p ) , so we only considerthe stationary points ˙ f ( p ) = 0 for p ∈ [1 , t − ] . In thefollowing, we will prove that there is no stationary pointfor p ∈ [1 , t − ] except for p = 1 .The equation (4) amounts to p ) = ( p − p ) log(2) , todiscuss whether this equation has solutions within (1 , t − ] ,we let h ( p ) = 2 log( p ) − ( p − p ) log(2) . (6)So we have ˙ h ( p ) = 2 p − (1 + 1 p ) log(2) = − ( 1 p − t )( 1 p − t ) . (7)Combining (6) and (7) tells that: 1) h ( p ) = 0 and ˙ h ( p ) = 0 when p = 1 ; 2) h ( p ) = 0 for p ∈ (1 , t − ] , because h ( p ) keepsdecreasing from h (1) = 0 as ˙ h ( p ) keeps negative when p changes from to t − . So ˙ f ( p ) = 0 does not have solutionswithin interval [1 , t − ] except for p = 1 . In addition, from(3) we have ¨ f (1) = log (2) − log(2) < . Based on Lemma1, 2 and 3, it is concluded that f ( p ) takes the maximal valueat p = 1 , so we have f ( p ) = ( 12 ) p + ( 12 ) p ≤ f (1) = 1 . (8)This theorem is proved. (cid:4) Theorem 2.
For p ≥ and ≤ x ≤ , ( x ) p + (1 − x ) p ≤ . Proof : Let f ( x ) = x p + (1 − x ) p . We have ˙ f ( x ) = px p − − p (1 − x ) p − (9)and ¨ f ( x ) = p ( p − x p − + 1 p ( 1 p − − x ) p − . (10)From derivative test, we know that the maximal valueof f ( x ) possibly lie at the end points of [0 , ] or at thestationary points within this interval. So we have two cases: 1) if ˙ f ( x ) = 0 for all x ∈ [0 , ] , then the maximal value of f ( x ) only lies at the end points, so f ( x ) = x p + (1 − x ) p ≤ max( f (0) , f ( 12 ))= max(1 , ( 12 ) p + ( 12 ) p ) . (11)2) if there are stationary points within [0 , ] , there is ˙ f ( x ) =0 , so based on (9) we have px p − = 1 p (1 − x ) p − ≡ C > . (12)At the stationary points, substituting (12) into (10) we have ¨ f ( x ) = C ( p − (cid:18) x − p (1 − x ) (cid:19) . (13)Because x ∈ [0 , ] and p ≥ , based on (13) there are ¨ f ( x ) (cid:26) = 0 , when p = 1 , x = ; > , otherwise. (14)In terms of Lemma 3, (14) tells that: 1) when ¨ f ( x ) = 0 ,the evaluations, p = 1 and x = , make f ( x ) = 1 .2) When ¨ f ( x ) > , the stationary points resulted from(12) corresponds to the local minimal extrema. That is tosay, in this case the maximal values of f ( x ) only lie atthe endpoints of [0 , ] , because f ( x ) is continuous and itsmaximal values cannot take at the stationary points wherethe twice derivative of f ( x ) is positive. Combining the twocases tells that the maximal value of f ( x ) is at the endpointsof [0 , ] or is , which has also been described by (11). Basedon Theorem 1, (11) means this theorem. (cid:4) Theorem 2 has many variants, such as the followings.
Remark 1. ( n ) p + (1 − n ) p ≤ , for p ≥ and n ∈ Z and n ≥ . Remark 2. ( − x ) p + ( x ) p ≤ , for p ≥ and ≥ x ≥ . Remark 3. sin p ( α ) + cos p ( α ) ≤ , for p ≥ and α ∈ [0 , π ] . In [6], a generalized H¨older inequality is introduced asfollows ( a λ a λ + a λ a λ )2 min(0 ,λ + λ − ≤ ( a + a ) λ ( a + a ) λ , a ij > , λ j > . If λ and λ take p and p respectively, no matter howto evaluate a ij , Theorem 2 cannot be derived from aboveinequality. If we forcibly make the following relation hold: a λ a λ = x p , a λ a λ = (1 − x ) p , a + a = 1 and a + a = 1 , it is too difficult to work out a ij , λ i becausethe equation set includes transcendental functions. So usingthe generalized H¨older inequality cannot derive Theorem2 directly. When λ and λ take identical values, H¨olderinequality degenerates into Cauchy-Schwarz equality, fromwhich Theorem 2 cannot be inferred too.Young inequality states that a λ b − λ ≤ λa + (1 − λ ) b for a, b > and > λ > , whose left side onlyhas a single term, so from Young inequality Theorem2 cannot be gotten. Minkowski inequality means that ( P k | a k + b k | p ) p ≤ ( P k | a k | p ) p + ( P k | b k | p ) p for p ≤ . It is apparent that Minkowski inequality cannot infer The-orem 2. Ky Fan inequality is still a hot spot, which states Q nk =1 a k ( P nk =1 a k ) n ≤ Q nk =1 (1 − a k )( P nk =1 (1 − a k )) n holds when < x ≤ ,apparently this inequality is also essentially different fromTheorem 2.The reference [3] has more than 5000 inequalities,wherein two inequalities seem close to Theorem 2, whichare ( 1 − x q + ( 1 + x q ≤ ( 1 + x p p − (15) s.t. x ∈ [0 , , p ∈ (1 , , p + 1 q = 1 and ( 1 − x p + ( 1 + x p ≤
12 (1 + x p ) , x ∈ [0 , , p ∈ (1 , . (16)Comparing (15) and (16) with Remark 2, we can see theyare actually different, and the left side of Remark 2 is largerthan that of (15) and (16) due to ( x ) p ≥ ( x ) p for p ≥ .So (15) and (16) are also completely different from Theorem2. Based on above comparisons, we think Theorem 2 is anew inequality which is independent upon known com-pared inequalities. The inequality combines two numberpairs: one pair composes of x and − x , and the otherpair p and p . It is very interesting that we can find thesummation of the first pair is , while the product of thesecond pair is . When the two pairs intertwine as shownin Theorem 2, the upper bound is . The two number pairsand the constant are pretty elegant, so we take it forgranted that the inequality shown in Theorem 2 is elegant.Theorem 1 is a foundation for proving Theorem 2, seeing p as the variable. The associated discussion happens in thearea corresponding to the 2nd pair. Theorem 1 sees x as thevariable, analysis is performed in the area correspondingto the 1st pair. So, in proving the proposed inequality, twoareas associated with the two pairs have been considered.In proving Theorem 1 and 2, only derivative test methodhas been used. However, contrast to the traditional meth-ods, in using the 1st derivative test to localize the extremepoints, we do not explicitly solve the points because thecorresponding equation are transcendental, and cannot beexplicitly solved. So, we straightforwardly use the equationof the 1st derivative test to discuss whether the extreme islocal maximal or minimal. On the other hand, we also usethe 1st derivative as well as the boundary condition of afunction to determine whether the function will get to within a special interval. These techniques, we think, canbe useful to seek and prove new inequalities consisting oftranscendental terms. ONCLUSIONS
An elegant inequality ( x ) p + (1 − x ) p < for p ≥ and ≥ x ≥ has been found and proved, whichcombines two number pairs, x and − x as well as p and p . It is fascinating that the summation and product of thepairs are . The inequality cannot be deduced from anycompared known famous inequalities. It has transcendentalterm, so some new techniques have been built to prove theinequality. Several variants of the inequality have also beengiven, it is anticipated that they are useful. A CKNOWLEDGMENT
This work is supported by NSFC under grants 61860206007and U19A2071, as well as the funding from Sichuan Uni-versity under grant 2020SCUNG205 R EFERENCES [1] J. Liu,
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