On Some Integral Means
aa r X i v : . [ s t a t . O T ] J a n On Some Integral means
Fariba Khoshnasib-Zeinabad and Mohammadhossein Mehrabi
Abstract.
Harmonic, Geometric, Arithmetic, Heronian and Contra-harmonic means have been studied by many mathematicians. In 2003, H. Evesstudied these means from geometrical point of view and established some ofthe inequalities between them in using a circle and its radius. In 1961, E.Beckenback and R. Bellman introduced several inequalities corresponding tomeans. In this paper, we will introduce the concept of mean functions andintegral means and give bounds on some of these mean functions and integralmeans.
1. Introduction
In their book of inequalities, Beckenback and Bellman established several in-equalities between arithmetic, harmonic and contra-harmonic means[ ]. Thesemeans are defined in the following paragraph, based on the original text by Eve[ ].Let a, b > a = b. Putting together the results from works of several math-ematicians, in particular Taneja established that max { a, b } > C > r > g > A >Hn > G > H > min { a, b } in [ ] and [ ], where C = a + b a + b is contraharmonicmean, r = q a + b is root square mean, g = a + ab + b )3( a + b ) is gravitational mean (alsocalled centroidal mean), A = a + b is arithmetic mean, Hn = a + √ ab + b is Heronianmean, G = √ ab is geometric mean and H = aba + b is harmonic mean of a and b .In this paper we introduce the notion of a mean function and utilize it to define someintegral means of a and b and then we establish some inequalities corresponding tothose mean functions and integral means.
2. Definitions and Main Theorems
All the means that appear in this paper are functions F with conditions a andb satisified:a) F : R → R + , where min { x, y } F ( x, y ) max { x, y } , Provided that ( x, y ) ∈ R , Mathematics Subject Classification.
Key words and phrases.
Mean function, integral mean, harmonic mean, arithmeticcomplimentary.Communicated by ... b) F ( x, y ) = F ( y, x ) , such that ( x, y ) ∈ R . Consequently, F ( x, x ) = x where x ∈ R + .We say F is a mean function when the two above conditions are satisfied.All throughout the paper we are assuming that a, b > b > a , by symmetry. Definition . Let M be a mean of a and b . We define M A := 2 A − M. tobe A -complementary (arithmetic complementary) of M . It is obvious that M A and M G are means of a and b . Theorem . Let M ∈ R ( R ) be a mean function. Then(i) I M := I M ( a, b ) := ( b − a ) R ba R ba M ( x, y ) dxdy, if a = b,a, if a = b is a mean of a and b ,(ii) J M := J M ( a, b ) := 3 I M ( a, b ) − A ( a, b ) is a mean of a and b and finally(iii) A < I M < A. Proof.
Let b > a .(i):min { a, b } = a < a + b b − a ) Z ba Z xa y dydx + 1( b − a ) Z ba Z bx x dydx =1( b − a ) Z ba Z ba min { x, y } dxdy I M ( a, b ) b − a ) Z ba Z ba max { x, y } dxdy = 1( b − a ) Z ba Z xa x dydx + 1( b − a ) Z ba Z bx y dydx = a + 2 b < b = max { a, b } . Also, it is obvious that I M is symmetric.(ii):By proof of (i), we have a + b I M ( a, b ) a +2 b .So,a I M ( a, b ) − ( a + b ) b. ( iii ) : 23 < a + b )3( a + b ) I M ( a, b ) A ( a, b ) a + 2 b )3( a + b ) < , multiplying by A ( a, b ) we get the result. (cid:3) N SOME INTEGRAL MEANS 3
Proposition . Let
M, M and M be mean functions and λ ∈ R . Then(i) M > M ⇒ I M ( a, b ) > I M ( a, b ) and J M ( a, b ) > J M ( a, b ) , a = b, (ii) I λM +(1 − λ ) M = λ I M + (1 − λ ) I M , if λM + (1 − λ ) M is a mean func-tion.In particular, I M A = ( I M ) A and I J M = J I M . Proof.
Proof is easily done by straightforward calculations. (cid:3)
Here are some examples where the above proposition is used:Let a = b. Example . I A ( a, b ) = 1( b − a ) Z ba Z ba
12 ( x + y ) dxdy = 12( b − a ) Z ba (cid:20) x xy (cid:21) ba dy = 14( b − a ) (cid:2) y ( b − a ) + y ( b − a ) (cid:3) ba = b + a A ( a, b ) . Example . I G ( a, b ) = 1( b − a ) Z ba Z ba √ xy dxdy = (cid:18) b − a Z ba √ t dt (cid:19) = (cid:18) b − a )3( b − a ) (cid:19) = g ( √ a, √ b ) . Example . I H ( a, b ) = 2( b − a ) Z ba Z ba xyx + y dxdy = 2( b − a ) Z ba (cid:2) xy − y ln( x + y ) (cid:3) ba dy = 23( b − a ) (cid:2) y ( b − a ) + y ( b − a ) − ( y + b ) ln( y + b ) + ( y + a ) ln( y + a ) (cid:3) ba =43 (cid:18) A ( a, b ) + 1( b − a ) (cid:18) a ln A ( a, b ) a + b ln A ( a, b ) b (cid:19)(cid:19) , a = b. Example . Let a < b , then: √ b − a ) I r ( a, b ) = Z ba Z ba p x + y dxdy = Z π − ab Z b cos θa sin θ ρ dρdθ + Z tan − baπ Z b sin θa cos θ ρ dρdθ. Double integrals in the above expression are easily calculated and the final resultis: I r ( a, b ) = 13 √ b − a ) (cid:18) ( √ √ a + b ) − a ln ( b + √ a + b a ) − ... KHOSHNASIB-ZEINABAD AND MEHRABI b ln ( a + √ a + b b ) − ab p a + b (cid:19) , a = b. Similarly, by Proposition 2.1 (ii), we will have:
Example . Let a = b , then I C ( a, b ) = ( I H ) A ( a, b ) = − (cid:18) A ( a, b ) + 2( b − a ) (cid:18) a ln A ( a, b ) a + b ln A ( a, b ) b (cid:19)(cid:19) Example . Let a = b , then I g ( a, b ) = 43 I A ( a, b ) − I H ( a, b ) =49 (cid:18) A ( a, b ) − b − a ) (cid:18) a ln A ( a, b ) a + b ln A ( a, b ) b (cid:19)(cid:19) . Example . I Hn ( a, b ) = 23 I A ( a, b ) + 13 I G ( a, b ) = 13 (2 A ( a, b ) + g ( √ a, √ b ))= 2(13 A + 13 AG + G )27( A + G )and finally, Example . I A + G ( a, b ) = 12 I A ( a, b ) + 12 I G ( a, b ) = 12 ( A ( a, b ) + g ( √ a, √ b ))= 17 A + 17 AG + 2 G A + G ) . Because I A = J A = A, By Proposition 2.1 (i), we can propose the following:
Proposition . Let M be a mean function and a = b , then(i) M > ( < ) A ⇒ I M > ( < ) A, (ii) M > ( < ) A ⇒ J M > ( < ) A, (iii) I C > I r > I g > A > I Hn > I G > I H and J C > J r > J g > A > J Hn > J G > J H . By proposition 2.1 (ii) and proposition 2.2(iii), we infer I r ( a, b ) > A ( a, b ) − I H ( a, b ) > A ( a, b ) , a = b. N SOME INTEGRAL MEANS 5
Therefore, 3 I r ( a, b ) − A ( a, b ) > A ( a, b ) − I H ( a, b ) > A ( a, b ) , a = b. In other words, J r ( a, b ) > I C ( a, b ) > A ( a, b ) , a = b. Similarly, A ( a, b ) > I G ( a, b ) > J G ( a, b ) , a = b, I C ( a, b ) + I G ( a, b ) > A ( a, b ) > I r ( a, b ) + I H ( a, b ) , a = b and J C ( a, b ) + J G ( a, b ) > A ( a, b ) > J r ( a, b ) + J H ( a, b ) , a = b. Proposition . If a = b , then(i) A < I G < A ,(ii) − ln 2)3 A < I H < A ,(iii) A < I C < − A ,(iv) A < I Hn < A ,(v) A < I g < A ,(vi) A < I r < √ √ A, (vii) A < J G < A ,(viii) − A < J H < A ,(ix) A < J C < − A ,(x) A < J Hn < A ,(xi) A < J g < − A ,(xii) A < J r < √
2( ln(1 + √ A, where , , − ln 2)3 , − , , , √ √ , , − , − , and √ √ are the best possible bounds we found for theinequalities between the integral means and the mean functions. KHOSHNASIB-ZEINABAD AND MEHRABI
Proof. (i): If we take b = at , t > , then the following will be concluded: f ( t ) := t + t +1) t +1)( t +1) = I G ( a,b ) A ( a,b ) . Taking the derivative, we get: f ′ ( t ) = − t ( t − t +1) ( t +1) < . Therefore, f is strictlydecreasing. So, lim t →∞ f ( t ) = < f ( t ) < lim t → + f ( t ) = 1 , t > . (ii): If we take b = at, t > , then we will have f ( t ) := (1+ ( t +1) ln t +12 − t ln t ( t − ( t +1) ) = I H ( a,b ) A ( a,b ) .f ′ ( t ) = − f ( t )3( t − t − t +1) , where f ( t ) := ( t + 1) ln t +12 − ( t + 3 t ) ln t + t − t − t + 1 .f ′ ( t ) = 3( t + 1) ln t +12 − t + 2 t ) ln t + 3 t − t. f ′′ ( t ) = 6( t + 1) ln t +12 − t +1) ln t + 6 t − . f ′′′ ( t ) = 6 ln t +12 − t + 6 − t . f ′′′′ ( t ) = t ( t +1) > . Therefore, f ′′′ is strictly increasing, so f ′′′ ( t ) > lim t → + f ′′′ ( t ) = 0 , t > . Consequently, f ′′ isstrictly increasing, hence f ′′ ( t ) > lim t → + f ′′ ( t ) = 0 , t > . Therefore, f ′ is strictlyincreasing, so f ′ ( t ) > lim t → + f ′ ( t ) = 0 , t > . Thus, f is strictly increasing, hence f ( t ) > lim t → + f ( t ) = 0 , t > . Consequently, f ′ ( t ) < , t > . Therefore, f isstrictly decreasing, so − ln 2)3 = lim t →∞ f ( t ) < f ( t ) < lim t → + f ( t ) = 1 , t > . (iii): a = b ⇒ − ln 2)3 < I H ( a,b ) A ( a,b ) < ⇒ − = 2 − − ln 2)3 > − I H ( a,b ) A ( a,b ) = I C ( a,b ) A ( a,b ) > − . (iv): a = b ⇒ A ( a, b ) < I G ( a, b ) < A ( a, b ) ⇒ A = A + A < A + I G ( a, b ) = I Hn ( a, b ) < A + A = A. (v): a = b ⇒ − ln 2)3 < I H ( a,b ) A ( a,b ) < ⇒ = − − ln 2)9 > − I H ( a,b )3 A ( a,b ) = I g ( a,b ) A ( a,b ) > − = 1 . (vi): f ( t ) := √ (cid:18) k ( t + 1) − t ln √ t t − ln( t + √ t ) − t √ t ( t + 1)( t − (cid:19) = I r ( a, b ) A ( a, b ) , where a = bt, < t < k := √ √ . The we have: f ′ ( t ) = √ f ( t )3( t +1) ( t − ,where f ( t ) := (3 t + 2 t + 3) p t + t ( t + 3) ln 1 + √ t t + ... (3 t + 1) ln( t + p t ) − k ( t + 1) . Therefore, f ′ ( t ) = (3 t + 6 t ) ln 1 + √ t t + 3 ln( t + p t ) + (9 t + 3) p t − k ( t + 1) ,f ′′ ( t ) = (6 t + 6) ln 1 + √ t t + (18 t + 6) √ t − k ( t + 1) ,f ′′′ ( t ) = 6 (cid:18) t − t + 4 t − t − t ( t + 1) + ln 1 + √ t t − k (cid:19) N SOME INTEGRAL MEANS 7 and f ′′′′ ( t ) = 6(1 − t + 8 t − t + t ) t ( t + 1) . Since t ∈ (0 , , so 1 − t + 8 t − t + t = (1 − t ) + (8 − t + t ) t > . Hence, f ′′′′ > , . Consequently, f ′′′ will be strictly increasing. Therefore, f ′′′ ( t ) < lim t → − f ′′′ ( t ) =6( √ √ − k ) = 0 , for 0 < t < . Thus, f ′′ is strictly decreasing.Hence, f ′′ ( t ) > lim t → − f ′′ ( t ) = 12 ln(1 + √
2) + 12 √ − k = 0 , for 0 < t < . So, f ′ will be strictly increasing. Therefore, f ′ ( t ) < lim t → − f ′ ( t ) = 12 ln(1 + √
2) + 12 √ − k = 0 , for 0 < t < . Hence, f is strictly decreasing. So, f ( t ) > lim t → − f ( t ) = 8 ln(1 + √
2) + 8 √ − k = 0 , on (0 ,
1) and consequently f ′ < , . Thus, f will be strictly decreasing 0n (0 , . Therefore, 1 = lim t → − Theorem . Let M ∈ R ( R ) be a mean function and ϕ : [ α , β ] → [0 , ∞ ) be an integrable function; that is, ϕ ∈ R ([ α , β ]) , where α , β ∈ R and α <β . Besides, ψ : R + → R + is a Lipschitz function with the constant of 1; that is | ψ ( x ) − ψ ( y ) | | x − y | , ( x, y ) ∈ R . Then S M,ϕ,ψ := S M,ϕ,ψ ( a, b ) , defined in the following way: S M,ϕ,ψ := − ¯ ϕ + A ( a, b ) − A ( ψ ( a ) , ψ ( b ))+ 1 β − α Z β α M ( ϕ ( t )+ ψ ( a ) , ϕ ( t )+ ψ ( b )) dt is a mean of a and b , where ¯ ϕ := 1 β − α Z β α ϕ ( t ) dt. Proof. min { a, b } = A ( a, b ) − | a − b | A ( a, b ) − | ψ ( a ) − ψ ( b ) | A ( a, b ) − A ( ψ ( a ) , ψ ( b )) + min { ψ ( a ) , ψ ( b ) } S M,ϕ,ψ A ( a, b ) − A ( ψ ( a ) , ψ ( b )) + max { ψ ( a ) , ψ ( b ) } = A ( a, b ) + | ψ ( a ) − ψ ( b ) | A ( a, b ) + | a − b | { a, b } . Also, it is obvious that S M,ϕ,ψ is symmetric. (cid:3) Remark . Since ϕ ∈ R ([ α , β ]) ⇔ ϕ ◦ η ∈ R ([0 , , where η ( t ) := ( β − α ) t + α . So, without loss of generality, we can assume α =0 and β = 1 . KHOSHNASIB-ZEINABAD AND MEHRABI Remark . If ψ ( a ) = ψ ( b ) , then S M,ϕ,ψ ( a, b ) = A ( a, b ) . Remark . If ϕ = c > ( c is a constant), then S M,c,ψ = S M,c,ψ ( a, b ) = − c + M ( c + ψ ( a ) , c + ψ ( b )) − A ( ψ ( a ) , ψ ( b )) + A ( a, b ) . Remark . ( S M,c,ψ ) A = ( S M,c,ψ ) A ( a, b ) = c − M ( c + ψ ( a ) , c + ψ ( b )) + A ( ψ ( a ) , ψ ( b )) + A ( a, b ) . Proposition . Let M, M and M be mean functions and λ ∈ R . Then(i) S A,ϕ,ψ = A, (ii) M > M ⇒ S M ,ϕ,ψ > S M ,ϕ,ψ , a = b, In particular, M > ( < ) A ⇒ S M,ϕ,ψ > ( < ) A, (iii) S λM +(1 − λ ) M ,ϕ,ψ = λ S M ,ϕ,ψ + (1 − λ ) S M ,ϕ,ψ , if λM + (1 − λ ) M is amean function.In particular, S M A ,ϕ,ψ = ( S M,ϕ,ψ ) A . Proof. It is straightforward by direct calculations. (cid:3) Example . Let M = G, ϕ ( t ) = c > c is a constant) and ˇ ψ ( t ) = t − sin t , then N c := N c ( a, b ) := S G,c, ˇ ψ ( a, b ) = A ( a, b ) − (cid:18) √ c + a − sin a − √ c + b − sin b (cid:19) . In particular, N = N ( a, b ) = A ( a, b ) − (cid:18) √ a − sin a − √ b − sin b (cid:19) . We can see( N c ) A = ( N c ) A ( a, b ) = A ( a, b ) + 14 (cid:18) √ c + a − sin a − √ c + b − sin b (cid:19) . In particular,( N ) A = ( N ) A ( a, b ) = A ( a, b ) + 14 (cid:18) √ a − sin a − √ b − sin b (cid:19) . Example . Let M = G, ϕ ( t ) = c > c is a constant) and ˜ ψ ( t ) =ln( t + 1) , then L c := L c ( a, b ) := S G,c, ˜ ψ ( a, b ) = − c + A ( a, b ) − ln( p ( a + 1)( b + 1)) + ... p ( c + ln( a + 1))( c + ln( b + 1)) . In particular, L = L ( a, b ) = A ( a, b ) − ln ( p ( a + 1)( b + 1)) + p (ln( a + 1))(ln( b + 1)) . N SOME INTEGRAL MEANS 9 Also we can see( L c ) A = ( L c ) A ( a, b ) = c + A ( a, b ) + ln( p ( a + 1)( b + 1)) − ... p ( c + ln( a + 1))( c + ln( b + 1)) . In particular,( L ) A = ( L ) A ( a, b ) = A ( a, b ) + ln( p ( a + 1)( b + 1)) − p (ln( a + 1))(ln( b + 1)) . Example . Let M = H, ϕ ( t ) = Id ( t ) = t, then J ψ := J ψ ( a, b ) := S H,Id,ψ ( a, b ) = A ( a, b ) − (cid:18) ψ ( a ) − ψ ( b )2 (cid:19) ln (cid:18) 1+ 1 A ( ψ ( a ) , ψ ( b )) (cid:19) . In particular,(2.1) J Id = J Id ( a, b ) = A ( a, b ) − (cid:18) a − b (cid:19) ln (cid:18) A ( a, b ) (cid:19) . We can see( J ψ ) A = ( J ψ ) A ( a, b ) = A ( a, b ) + (cid:18) ψ ( a ) − ψ ( b )2 (cid:19) ln (cid:18) A ( ψ ( a ) , ψ ( b )) (cid:19) . In particular,( J Id ) A = ( J Id ) A ( a, b ) = A ( a, b ) + (cid:18) a − b (cid:19) ln (cid:18) A ( a, b ) (cid:19) . Example . Let M = G and ψ ( t ) = Id ( t ) = t, then(2.2) I ϕ := I ϕ ( a, b ) := S G,ϕ,Id ( a, b ) = − ¯ ϕ + Z p ( ϕ ( t ) + a )( ϕ ( t ) + b ) dt. In particular,(2.3) I Id = I Id ( a, b ) = − 12 + F ( a + 1 , b + 1) − F ( a, b ) − F ( a + 1 , b + 1) + F ( a, b ) , where(2.4) F ( x, y ) := 14 ( x + y ) √ xy, F ( x, y ) := 14 ( x − y ) ln √ x + √ y √ . We can see( I ϕ ) A = ( I ϕ ) A ( a, b ) = a + b + ¯ ϕ − Z p ( ϕ ( t ) + a )( ϕ ( t ) + b ) dt. In particular,( I Id ) A = ( I Id ) A ( a, b ) = a + b + 12 − F ( a +1 , b +1)+ F ( a, b )+ F ( a +1 , b +1) − F ( a, b ) . If a = b, by proposition4 (ii), we will have N c < A, L c < A, J ψ < A, I ϕ < A. Proposition . If a = b, then I ϕ > G, for every ϕ, whose support ispositive measure; equivalently, there exists S ⊆ [0 , with | S | > , such that ϕ ( t ) > for every t ∈ S. Proof. Let a = b. We know that I ϕ > G is equivalent to Z (cid:18)p ( ϕ ( t ) + a )( ϕ ( t ) + b ) − ϕ ( t ) − G ( a, b ) (cid:19) dt > . Let us start our argument by working with the integrand: p ( ϕ ( t ) + a )( ϕ ( t ) + b ) − ϕ ( t ) − G ( a, b ) > ( > )0This results in: ( √ a − √ b ) ϕ ( t ) > ( > )0 . Thus, p ( ϕ ( t ) + a )( ϕ ( t ) + b ) − ϕ ( t ) − G ( a, b ) > , t ∈ [0 , 1] and p ( ϕ ( t ) + a )( ϕ ( t ) + b ) − ϕ ( t ) − G ( a, b ) > t ∈ S . By (2) and (2), we will have Z (cid:18)p ( ϕ ( t ) + a )( ϕ ( t ) + b ) − ϕ ( t ) − G ( a, b ) (cid:19) dt > . Hence, I ϕ > G. (cid:3) Proposition . If a = b, then(i) J Id > H, (ii) neither J Id > G nor J Id < G ,(iii) neither L < G nor L > H ,(iv) neither N < G nor N > H. Proof. Let a = b . (i): J Id > H ⇔ ( a − b ) a + b ) > ( a − b ) ln (1 + a + b ) ⇔ a + b > ln(1 + a + b ) X . (ii) One counter example is : J Id (0 . , < . < . < G (0 . , . On the other hand, J Id (0 . , . > . > . > G (0 . , . . (iii) A counter examples would be: L (0 . , . > . > . > G (0 . , . . On the other hand, we have: L (4 . , . − H (4 . , . < − − < . (iv) Here is a counter example: N (0 . , . > . > . > G (0 . , . N (4 . , . − H (4 . , . < − − < . (cid:3) N SOME INTEGRAL MEANS 11 Theorem . Let M ∈ R ( R ) be a mean function and ϕ : [ α , β ] → R + bean integrable function; that is, ϕ ∈ R ([ α , β ]) , where α , β ∈ R and α <β . Also, ψ : R + → R + is a Lipschitz function with the constant of 1.Then P M,ϕ,ψ := P M,ϕ,ψ ( a, b ) , defined in the following way: P M,ϕ,ψ ( a, b ) := A ( a, b ) − A ( ψ ( a ) , ψ ( b )) + 1( β − α ) ¯ ϕ Z β α M ( ϕ ( t ) ψ ( a ) , ϕ ( t ) ψ ( b )) dt is a mean of a and b . Proof. min { a, b } = A ( a, b ) − | a − b | A ( a, b ) − | ψ ( a ) − ψ ( b ) | A ( a, b ) − A ( ψ ( a ) , ψ ( b )) + min { ψ ( a ) , ψ ( b ) } P M,ϕ,ψ ( a, b ) ...A ( a, b ) − A ( ψ ( a ) , ψ ( b )) + max { ψ ( a ) , ψ ( b ) } = A ( a, b ) + | ψ ( a ) − ψ ( b ) | A ( a, b ) + | a − b | { a, b } . Also, it is obvious that P M,ϕ,ψ is symmetric. (cid:3) Remark . Since ϕ ∈ R ([ α , β ]) ⇔ ϕ ◦ η ∈ R ([0 , , where η ( t ) := ( β − α ) t + α . So, without loss of generality, we can assume α =0 and β = 1 . Remark . If ψ ( a ) = ψ ( b ) , then P M,ϕ,ψ ( a, b ) = A ( a, b ) . Remark . If ϕ = c > ( c is a constant), then P M,c,ψ = P M,c,ψ ( a, b ) = 1 c M ( cψ ( a ) , cψ ( b )) − A ( ψ ( a ) , ψ ( b )) + A ( a, b ) . Remark . ( P M,c,ψ ) A = ( P M,c,ψ ) A ( a, b ) = − c M ( cψ ( a ) , cψ ( b )) + A ( ψ ( a ) , ψ ( b )) + A ( a, b ) . Remark . If M is a homogeneous function of order 1; that is M ( xz, yz ) = zM ( x, y ) , ∀ x, y, z ∈ R + , then P M,ϕ,ψ ( a, b ) = A ( a, b ) − A ( ψ ( a ) , ψ ( b )) + M ( ψ ( a ) , ψ ( b )) . Remark . C, r, g, A, Hn, G and H are homogeneous mean functions oforder 1. Proposition . Let M, M and M be mean functions and λ ∈ R . Then(i) The following three equations hold: P A,ϕ,ψ ( a, b ) = A ( a, b ) , P G,ϕ,ψ ( a, b ) = A ( a, b ) − A ( ψ ( a ) , ψ ( b )) + G ( ψ ( a ) , ψ ( b )) = A ( a, b ) − ( p ψ ( a ) − p ψ ( b )) , ( P G,ϕ,ψ ) A ( a, b ) = A ( a, b ) + ( p ψ ( a ) − p ψ ( b )) , (ii) M > M ⇒ P M ,ϕ,ψ > P M ,ϕ,ψ , a = b, In particular, M > ( < ) A ⇒ P M,ϕ,ψ > ( < ) A, (iii) P λM +(1 − λ ) M ,ϕ,ψ = λ P M ,ϕ,ψ + (1 − λ ) P M ,ϕ,ψ , if λM + (1 − λ ) M is amean function.In particular, P M A ,ϕ,ψ = ( P M,ϕ,ψ ) A . Proof. It is straightforward. (cid:3) Here are some examples where the above proposition is used: Example . Let M = G and ˇ ψ ( t ) = t − sin t , then P G,ϕ, ˇ ψ = N . Example . Let M = G and ˜ ψ ( t ) = ln( t + 1) , then P G,ϕ, ˜ ψ = L . Example . Let M = H, ϕ ( t ) = Id ( t ) = t, then P H,Id,ψ ( a, b ) = A ( a, b ) − A ( ψ ( a ) , ψ ( b )) + H ( ψ ( a ) , ψ ( b )) = A ( a, b ) − ( ψ ( a ) − ψ ( b )) ψ ( a ) + ψ ( b )) . In particular, P H,Id,Id ( a, b ) = H ( a, b ) . We can see P H,Id,ψ ( a, b ) < S H,Id,ψ ( a, b ) , a = b. Let M be a mean function. We defineˆ S M := ˆ S M ( a, b ) := Z π M ( a sin θ, b cos θ ) dθ. We can easily see ˆ S M ( a, b ) = ˆ S M ( b, a ) . Also,ˆ S M ( a, b ) Z π ( a sin θ + b cos θ | a sin θ − b cos θ | dθ = A ( a, b ) + 12 Z tan − ba ( b cos θ − a sin θ ) dθ + 12 Z π tan − ba ( − b cos θ + a sin θ ) dθ = p a + b . ˆ S M ( a, b ) > Z π ( a sin θ + b cos θ − | a sin θ − b cos θ | dθ = A ( a, b ) − Z tan − ba ( b cos θ − a sin θ ) dθ − Z π tan − ba ( − b cos θ + a sin θ ) dθ =2 A ( a, b ) − p a + b . N SOME INTEGRAL MEANS 13 Thus, by (2) and (2), we have(2.5) 2 A ( a, b ) − p a + b ˆ S M p a + b . From (2.5), if we take ξ := ξ ( a, b ) > ζ := ζ ( a, b ) , such that(2.6) min { a, b } ( a + b ) ξ √ a + b − ξ + ζ ξ √ a + b ˆ S M + ζ ξ + ζ max { a, b } , then we will have(2.7) 0 < ξ | a − b |√ a + b √ a + b − ( a + b )and(2.8) min { a, b } + (1 − a + b √ a + b ) ξ ζ max { a, b } − ξ. By (2.8), (2.9) and (2.8), we infer(2.9) S M,ξ,ζ := S M,ξ,ζ ( a, b ) := ξ ( a, b ) √ a + b ˆ S M ( a, b ) + ζ ( a, b )is a mean of a and b .For example, if we take ξ := | a − b | and ζ := ( min { a, b } + (1 − a + b √ a + b ) | a − b | +max { a, b } − | a − b | ) = A ( a, b ) − | a − b | √ a + b , then from (2.9)(2.10) S M := S M ( a, b ) := A ( a, b ) − | a − b | √ a + b + | a − b |√ a + b Z π M ( a sin θ, b cos θ ) dθ is a mean of a and b . Thus, we will have the following theorem Theorem5 Let M is a mean function. Then S M which is defined by (2.10),is a mean function. Proposition . Let M, M and M be mean functions and λ ∈ R . Then(i) S λM +(1 − λ ) M = λS M + (1 − λ ) S M , if λM + (1 − λ ) M is a mean func-tion.Specially, S M A = ( S M ) A , (ii) M > M ⇒ S M > S M , a = b, specially, M > ( < ) A ⇒ S M > ( < ) A. Proof. is straightforward. (cid:3) Example . S A = A, S G ( a, b ) = A ( a, b ) + G ( a, b ) | a − b | Γ ( ) p π ( a + b ) − | a − b | √ a + b , Γ( ) ≈ . . Example . S H ( a, b ) = A ( a, b ) − | a − b | ( a + b − ab )2( a + b ) − ... a b | a − b | ( a + b ) ln a + b + √ a + b √ ab . Example . By proposition 2.8 (i) S g = A ( a, b ) + | a − b | ( a + b − ab )6( a + b ) + ... a b | a − b | a + b ) ln a + b + √ a + b √ ab and Example . S C = A ( a, b ) + | a − b | ( a + b − ab )2( a + b ) + ... a b | a − b | ( a + b ) ln a + b + √ a + b √ ab . By proposition2.8 (ii), for a = b (cid:18) p a + b ) | a − b | (cid:19) S C − a + b √ (cid:18) √ a + b | a − b | − (cid:19) > Z π p a sin θ + b cos θ dθ > (cid:18) p a + b ) | a − b | (cid:19) S g − a + b √ (cid:18) √ a + b | a − b | − (cid:19) and S C ( a, b ) > S g ( a, b ) > A ( a, b ) > S G ( a, b ) > S H ( a, b ) , a = b. Specially, A (3 , > S G (3 , > S H (3 , , which we infer712 > Γ ( ) √ π > − 48 ln6125 . Theorem . Let M , M ∈ R ( R ) be mean functions. Then T M ,M := T M ,M ( a, b ) := b − a R ba M ( M ( a, b ) , x ) dx, a = b,a, a = b is a mean of a and b . N SOME INTEGRAL MEANS 15 Proof. Let b > a and x ∈ [ a, b ] . We have1 b − a Z ba M ( M ( a, b ) , x ) dx b − a Z ba max { M ( a, b ) , x } dx =1 b − a Z M ( a,b ) a M ( a, b ) dx + 1 b − a Z bM ( a,b ) x dx = ... b − a ) ( b + M ( a, b ) − aM ( a, b )) . .If we take m ( t ) := b + t − at, t ∈ [ a, b ] , then m will be increasing on [ a, b ] . So, m ( t ) m ( b ) = 2 b ( b − a ) , for t ∈ [ a, b ] . Hence, from (2), we will get1 b − a Z ba M ( M ( a, b ) , x ) dx b. Similarly, 1 b − a Z ba M ( M ( a, b ) , x ) dx > b − a Z ba min { M ( a, b ) , x } dx =1 b − a Z M ( a,b ) a x dx + 1 b − a Z bM ( a,b ) M ( a, b ) dx = ... b − a ) ( − a − M ( a, b ) + 2 bM ( a, b )) .. If we take m ( t ) := − a − t + 2 bt, t ∈ [ a, b ] , then m will be increasing on [ a, b ] . So, m ( t ) > m ( a ) = 2 a ( b − a ) , for t ∈ [ a, b ] . Therefore, from (2), we will get1 b − a Z ba M ( M ( a, b ) , x ) dx > a. Also, it is obvious that T M ,M is symmetric. (cid:3) Proposition . Let M , M ′ , M and M ′ be mean functions and λ ∈ R .Then(i) M > M ′ ⇒ T M ,M ( a, b ) > T M ′ ,M ( a, b ) , a = b, (ii) If M is strictly increasing and M > M ′ , then T M ,M ( a, b ) > T M ,M ′ ( a, b ) , a = b, (iii) T λM +(1 − λ ) M ′ ,M = λ T M ,M + (1 − λ ) T M ′ ,M , if λM + (1 − λ ) M ′ is a mean function.Specially, T M A ,M = 2 T A,M − T M ,M . Proof. is straightforward. (cid:3) Some Examples Let M be a mean function.(1) T A,M ( a, b ) = A ( M ( a, b ) , A ( a, b )) . (2) T G,M ( a, b ) = G ( M ( a, b ) , g ( √ a, √ b )) . (3) T H,M ( a, b ) = 2 M ( a, b ) (cid:18) − M ( a, b ) b − a ln b + M ( a, b ) a + M ( a, b ) (cid:19) , a = b. (4) T r,M ( a, b ) = 12 √ (cid:18) ( a + b )( a + b + M ( a, b ))( a p a + M ( a, b ) + b p b + M ( a, b )) + M ( a, b )( b − a ) ln b + p b + M ( a, b ) a + p a + M ( a, b ) (cid:19) , a = b. By proposition 2.9, we will get T r,M ( a, b ) > T A,M ( a, b ) > T G,M ( a, b ) > T H,M ( a, b ) , a = b. Also, by proposition2.9(ii) and note12, we will have T A,A ( a, b ) > T A,G ( a, b ) > T A,H ( a, b ) , a = b, T G,A ( a, b ) > T G,G ( a, b ) > T G,H ( a, b ) , a = b and T H,A ( a, b ) > T H,G ( a, b ) > T H,H ( a, b ) , a = b. Besides, by proposition2.9(iii), we will have T H n ,M = T A,M + T G,M , T g,M = T A,M − T H,M and T ( M ) A ,M = 2 T A,M − T M ,M , if M, M and M are mean functions. References 1. H. Eves, Means Appearing in Geometrical Figures , Mathematics Magazine (2003), 292-294.2. E. Beckenback, R. Bellman, An Introduction to Inequalities , Random House, Inc., New York,1961.3. Inder J. Taneja, Refinement of Inequalities among Means , Journal of Combinatorics, Informa-tionand Systems Sciences (2006), 357-378.4. Inder J. Taneja, Inequalities having seven means and proportionality relations , arXiv:1203.2288 Department of Mathematics, Earlham College, Richmond, IN, United States E-mail address : [email protected] Department of Mathematics, Iran University of Science and Technology, Provinceof Tehran, Iran E-mail address ::