... =1+ x a+b c 24 x =1 a+b c 25 z =1 a b+c 10 /3 52/9 26 z=0 a b+c - 40/9 27 z =1 a b+c 28 x =1 a+b c 29 z =1 a b+c 10 /3 52/9 30 z=0 a b+c 18 /5 11 4/25 31 y =1 a+b c 14 /5 32 z=0 a b+c 18 /5 11 4/25 33 y =1 ... only 16 possibilities: A0 = 1, A1 = 1, B0 = 1 and B1 = 1 Consider the CHSH expression, CHSH = A0 B0 + A0 B1 + A1 B0 − A1 B1 , (2 .15 ) = A0 (B0 + B1 ) + A1 (B0 − B1 ) (2 .16 ) Note that (B0 + B1 ... = ax a+c b 12 z = ax a+c b 13 y =1+ z a b+c - 40/9 14 y =1+ x a+b c 10 /3 52/9 15 x=0 a+b c 16 y =1 a+b c - 40/9 17 x=0 a+b c 18 z=0 a b+c 19 z =1 a b+c 9/2 20 z=0 a b+c 16 /5 26/5 21 z =1 a b+c 9/2...
... r g Sk +1 h i f f IE 1 + r,k +1 gSk +1 jF k = 1 + r,k IE + r g Sk +1 jF k f 1 + r,k IE g + r Sk +1 jF k f 1 + r,k g IE + r Sk+1jF k = 1 + r,k g Sk ; So f 1 + r,k ... 0; 1; : : : ; n: Example 7 .1 (Stopped Process) Figure 7.3 shows our familiar 3-period binomial example Define ! = Then !1 = T; !1 = H: S HH = 16 S2^ ! ! = if if : S2 HT = S1 T = S1 ... f0; 1; : : : ; ng k The set f kg is in F k , so the set f k + 1g = f kgc is also in F k We compute h i h IE Yk +1 ^ jF k = IE If kgY + If k+1gYk +1 jF k = If kgY + If k+1gIE Yk +1 jF...
... 2 011 , 2 011 : 71 http://www.journalofinequalitiesandapplications.com/content/2 011 /1/ 71 Page of 1/ s s(p−t)/(s−t) 1/ s (f (x)+g(x)) dx ≥ p s f (x)dx s + g (x)dx × f t (x)dx (1: 4) 1/ t t(s−p)/(s−t) 1/ t ... Inequalities and Applications 2 011 , 2 011 : 71 http://www.journalofinequalitiesandapplications.com/content/2 011 /1/ 71 ˜ ˜ ˜ Wi (K +L )1/ (n−i) ≤ Wi (K )1/ (n−i) + Wi (L )1/ (n−i) , ˜ with equality if and ... doi :10 .10 16/00 01- 8708(88)90077 -1 Gardner, RJ: Geometric Tomography Cambridge University Press, New York (19 96) Lutwak, E: Dual mixed volumes Pacific J Math 58, 5 31 538 (19 75) doi :10 .11 86 /10 29-242X-2 011 -71...
... m n m r =1 j =1 Arj (1 − er + es ) Asi s =1 i =1 m s =1 i =1 r =1 j =1 n m m n s =1 i =1 j =1 r =1 = n λj Arj (1 − er + es ) n m Asi ≥ λ Arjj (1 − er Asi n = A 1 s1 s =1 n λj + es ) − 11 Ar1 (1 − er + ... n m n m r =1 j =1 Arj (1 − er + es ) Asi s =1 i =1 n n − 1 − er + e s ) A 1 A 1 (1 s1 r1 ≥ λj m j=2 s =1 r =1 m n λ A 1 Arjj (1 s1 × m n n × s =1 r =1 j=2 − 1 Aλ r1 = λj − er + e s ) s =1 r =1 j=2 n λj ... Asjj A 1 (1 r1 n m j=2 λj × m r =1 n n × λ Asjj A 1 (1 r1 λ A 1 Arjj (1 − er + es ) s1 s =1 r =1 j=2 n n λj − er + e s ) ( 21) s =1 r =1 n − 1 Aλ r1 = m j=2 λj × m r =1 Aλ s1 − s =1 n λ Arjj er + r =1 n...
... http://www.journalofinequalitiesandapplications.com/content/2 011 /1/ 48 Page of 15 and hence ψ(ϕ 1 ( (1 − t)u + tv)) ≤ (1 − t)ψ(ϕ 1 (u)) + tψ(ϕ 1 (v)) for all t Î (0, 1) This means that ψ ∘ -1 is convex Conversely, if ψ ∘ -1 is convex, we ... λc,ξs1 ,ξs2 (x) = s2 λc,ϕ,ψ (x) + − s2 (s1 = 0) (3) and λc,ξs1 ,ξs2 (x) = s2 s − s1 − s1 s1 λc,ϕ,ψ (x) − s1 1 s1 (s1 = 0) (4) for each x Î I\{c} If s1 = 0, then it is trivial by (3) that λc,ξs1 ... Applications 2 011 , 2 011 :48 http://www.journalofinequalitiesandapplications.com/content/2 011 /1/ 48 Page of 15 Then u Î (I)\{d} and d Î ((I))∘, hence (λc,ϕ,ψ ◦ ϕ 1 )(u) = (ψ ◦ ϕ 1 )(u) − (ψ ◦ ϕ 1 )(d)...
... let p1 , p2 , , pn be positive real numbers such that p1 p1 a1 p2 a2 p ··· p pn an p a 11 a22 · · · ann ≥ Qi ai , ak ··· 2. 21 ak 1 p2 , Q i aRk k 2.22 with equality for a1 Remark 2 .11 For p1 ··· ... 3 .14 where ··· pn xn pk 1 xk 1 Rk xk , pk xk Yk pk xk Rk p1 x1 Xk ··· X p1 x1 p2 x2 ··· , 3 .15 pn xn The inequality 3 .14 follows from Lemma 2.2, since xk , X ∈ Yk , Xk and Rk Xk Yk Rk xk X 3 .16 ... p1 p1 a1 p2 a2 ≥ Qi ··· p p p2 ··· 2 .10 pn Then, p pn an − a 11 a22 · · · ann Rk ak − Qi Qi / Qi Rk Rk R / Qi Rk ak k 2 .11 , Journal of Inequalities and Applications with equality for ··· a2 a1...
... page φ y2 − φ y1 φ x2 − φ x1 ≤ x2 − x1 y2 − y1 3. 21 with x1 ≤ y1 , x2 ≤ y2 , x1 / x2 , y1 / y2 Since by Theorem 3.6, Γt is log-convex, we can set in 3. 21 : φ x log Γx , x1 t, x2 s, y1 u, and y2 ... exp M1 ,1 exp 2u − u u 1 , log xi u 1 Pn Mu 1 x u / 0, 1, Pn log2 d − Pn M2 log x 2Pn log M0 x − 2d 2Pn log d − log M0 x Pn d log2 d n u 1 i pi xi u 1 duMu 1 x u P n Mu x − d u n i n i 1 − dM 1 ... results can also be given for 2 .10 and 2 .15 Namely, suppose that φ /ψ has inverse function, then from 2 .10 and 2 .15 we have ξ φ ψ ξ φ ψ 11 φ d ψ d 1/ Pn 1/ Pn n i pi n i pi xi − d φ xi − 1/ Pn...
... no 4, pp 422–423, 19 69 [12 ] J P Williams, “Problems and solutions: solutions of advanced problems: 5642,” The American Mathematical Monthly, vol 76, no 10 , pp 11 53 11 54, 19 69 [13 ] J.-L Li, “On ... n +1 π − (2x) (n ≥ 2) x π π (3 .13 ) (3 .14 ) hold with equality if and only if x = π/2 Furthermore, the constants (π − 2)/π and 2/π in (3 .13 ), as well as the constants 2/(nπ n +1 ) and (π − 2)/π n +1 ... (2 .13 ) is slightly stronger than inequality (2 .15 ) In the case when x ∈ (0,x0 ], inequality (2 .13 ) is slightly stronger than inequality (1. 2), where 7π + 240π − 2880 x0 = 12 π 1/ 2 ≈ 1. 2 (2 .16 )...
... k 1/ 2 T n dt 1/ 2 x t − nr ˙ dt T n 1 x t ˙ dt n2 κ2 T x t ˙ dt, 2 .10 that is, ˙ ¨ ⇒ T x ≤ nκT x ˙ x ≤ nκ x ¨ From 2 .1 and T |x ˙ t |2 dt 2 .11 0, we have ¨ ˙ 2π x ≤ T x 2 .12 Combining 2 .11 and ... inequality, one may see 15 and the references therein Now, we are ready to prove our main results We first give the proof of Theorem 1. 3 Proof of Theorem 1. 3 From 1.1 and Definition 1. 1, for all t, u ∈ ... Applications, vol 14 , no 8, pp 7 01 715 , 19 90 T Furumochi, “Existence of periodic solutions of one-dimensional differential-delay equations,” Tohoku Mathematical Journal, vol 30, no 1, pp 13 –35, 19 78 S Chapin,...
... > 1, 1/ p 1/ q 1, r > 1, 1/ r 1/ s 1, t ∈ 0, , − min{r, s} t min{r, s} ≥ λ > p q − min{r, s} t, such that ∞ > ∞ np 1 t 2t−λ /r 1 an > 0, ∞ > ∞ nq 1 t 2t−λ /s 1 bn > 0, n n then ∞ ∞ am bn m n 1m ... we have K o1 K o1 O
... assume that x1 ∈ 1, and x2 ∈ 0, The equation of the straight line through points x1 , f x1 , 0, is y Since f is convex on 1, and x1 < p1 x1 f p1 x1 p1 p2 x2 p2 f x1 x x1 p2 x2 / p1 ≤ 2 .1 p2 ≤ 0, ... f x1 p1 x1 x1 p1 p2 x2 p2 2.2 S Hussain et al It is enough to prove that f x1 p1 x1 x1 p1 p2 x2 p1 f x1 ≤ p2 p1 p2 f x2 p2 2.3 which is obviously equivalent to the inequality f x1 f x2 ≤ x1 x2 ... , xn 1 i xi − xi , i 1, , n If xi xn 1 i ≤ and A2 xi , xn Gn Gn ≥ n i A2 xi , xn A2 xi , xn 1 i pi pn 1 i 1 i pi xi pn 1 i xn 1 i / pi pn ≤ 1/ 2, i 1, , n, then 1 i 1/ 2Pn ≥ An An 1 i ,...
... k 1g 2N 2k 1 1/ 2N 1 α2k g 2k ln N k α2k 2k ln g t ∇t 1 α 2k ln g 2k 2N − N 1 k g 2N − ln 1 α 2N ln 1 α 2k 1 α 2k k g 1/ 2N 1 k N 1 ln ⎝ 3.20 1 α 2k N k ⎛ g 2k 1/ 2N 1 k N α2 g 2k k g 2k 1 α ... 1 α 2k 1/ 2N 1 ⎞ ⎠ k We conclude that ln α N 1 k k g ⎛ ≥ ln ⎝ 2k N 1 1−α 2N − k N k k 12 g 2k 1/ 2N 1 N α2 g 2k k g 2k k 1/ 2N 1 1−α ⎞ 3. 21 ⎠ k On the other hand, N 1 α 2k g 2k k 1 α N N 1 2k ... 2N ln α 1 ≥ 2N 2N g 2N 1 α 1 N 1 k k g 2N − α ln t Δt k t ∇t 1 N k k 12 g 2N − 1 α 1, b 2N , and g : {2k : k ln g t ♦α t 2N − α α 2N ln g t Δt 2N 1 N 1 k k ln g 2N − N 1 k α2k N 1 ln 2N −...
... tx1 f q,t,x1 qt x1 t,x1 θ∈ 0 ,1/ 1 q tx1 x− σ2 − q tx1 x1 qtx1 − t x1 , 2b 1 2.8 1 − qtx1 q − q t2 x1 − q tx1 21 1 − qtx1 x − θσ − qtx1 Maximizing over x1 , we get 1 t x1 −f x − θσ θ∈ 0 ,1/ 1 q ... 1 α 2x1 qtx1 − qtx1 1 α 2x1 1 α 1 α 2x α qtx1 1 α 1111 α α− Ye Xia 13 Maximizing the last expression over q, we get − α / 2x1 Maximizing the result over x1 , we get 1 α 2a 1 2 .10 Also, ... x − θσ θ∈ 0 ,1/ 1 q tx1 θ∈ 0 ,1/ 1 q tx1 −f x − θσ − qtx1 x − θσ − qtx1 k 2x1 qtx1 − qtx1 k 2x1 k − qtx1 k k 1111 Maximizing the last expression over q and t, we get 1/ k x1 / k −bk/ k...
... Δu f 1/ v−u 2 .12 Journal of Inequalities and Applications Proof An equivalent form of 1. 12 is ϕ y2 − ϕ y1 ϕ x2 − ϕ x1 ≤ , x2 − x1 y2 − y1 2 .13 where x1 ≤ y1 , x2 ≤ y2 , x1 / x2 , and y1 / y2 ... Theorem 1.1 Let f and p be nonnegative and integrable functions on a, b , with then for < r < s < t, r, s, t / 1, one has t−r Ds s s 1 ≤ t−s Dr r r 1 Dt t t 1 b p a x dx 1, s−r 1. 3 Remark 1. 2 For ... s is log-convex, and Γα s s s 1 Γα α s α sB s 1, α s − f t tα dt log f t tα dt Γα 1 α 1 H α 1 − H α log α α log α 2 f s t tα dt , s / 0, 1, 1 α 1 α f t tα dt , 3 .16 f t tα dt f t tα dt log f t...
... x 1 ε x 1 ε ∞ ∞ t ( 1 ε)/2 dt dx 1/ x Amin {1, t } + B max {1, t } ∞ x 1 ε (2 .13 ) t ( 1 ε)/2 dt dx Amin {1, t } + B max {1, t } 1/ x t ( 1 ε)/2 dt dx Amin {1, t } + B max {1, t } For x ≥ 1, we get 1/ x ... ( 1 ε)/2 dt Amin {1, t } + B max {1, t } 1/ x ( 1 ε)/2 t At + B 1/ x dt ≤ B = 1 B − (1 + ε)/2 x ≤ 1/ 4 x B t ( 1 ε)/2 dt (2 .14 ) 1 (1+ ε)/2 (setting < ε < 1/ 2) Thus ∞ 0< ≤ B ≤ B x 1 ε ∞ ∞ 1/ x t ( 1 ε)/2 ... y } x ∞ 1/ 2 dt + t At + B A 1+ (1/ 2) AB B 1/ 2 √ AB =√ (setting t = 1/ u) dx t 1/ 2 dt Amin {1, t } + B max {1, t } 1/ 2 A/B ∞ A/B 1/ 2 dt t A + Bt 1 1/ 2 dt + √ t 1+ t AB 1 1/ 2 u du + √ 1+ u AB ∞...
... x 1 ε x 1 ε ∞ ∞ t ( 1 ε)/2 dt dx 1/ x Amin {1, t } + B max {1, t } ∞ x 1 ε (2 .13 ) t ( 1 ε)/2 dt dx Amin {1, t } + B max {1, t } 1/ x t ( 1 ε)/2 dt dx Amin {1, t } + B max {1, t } For x ≥ 1, we get 1/ x ... ( 1 ε)/2 dt Amin {1, t } + B max {1, t } 1/ x ( 1 ε)/2 t At + B 1/ x dt ≤ B = 1 B − (1 + ε)/2 x ≤ 1/ 4 x B t ( 1 ε)/2 dt (2 .14 ) 1 (1+ ε)/2 (setting < ε < 1/ 2) Thus ∞ 0< ≤ B ≤ B x 1 ε ∞ ∞ 1/ x t ( 1 ε)/2 ... y } x ∞ 1/ 2 dt + t At + B A 1+ (1/ 2) AB B 1/ 2 √ AB =√ (setting t = 1/ u) dx t 1/ 2 dt Amin {1, t } + B max {1, t } 1/ 2 A/B ∞ A/B 1/ 2 dt t A + Bt 1 1/ 2 dt + √ t 1+ t AB 1 1/ 2 u du + √ 1+ u AB ∞...
... obtain ∞ λn +1 a 1 aλ2 · · · aλn n 1/ Λn ∞ = n =1 λn +1 λ 1/ Λn λ λ c 11 c22 · · · cnn n =1 ∞ λn +1 λ λ λ c 11 c22 · · · cnn ≤ n =1 ∞ = λm cm am m =1 ∞ c1 a1 1 c2 a2 λ2 · · · cn an λn 1/ Λn n 1/ Λn λm cm ... results ∞ n =1 an Theorem 1. 3 (see [3, Theorem 1] ) Let an ≥ 0(n ∈ N) and < ∞ 1/ n a1 a2 · · · a n
... obtain ∞ λn +1 a 1 aλ2 · · · aλn n 1/ Λn ∞ = n =1 λn +1 λ 1/ Λn λ λ c 11 c22 · · · cnn n =1 ∞ λn +1 λ λ λ c 11 c22 · · · cnn ≤ n =1 ∞ = λm cm am m =1 ∞ c1 a1 1 c2 a2 λ2 · · · cn an λn 1/ Λn n 1/ Λn λm cm ... results ∞ n =1 an Theorem 1. 3 (see [3, Theorem 1] ) Let an ≥ 0(n ∈ N) and < ∞ 1/ n a1 a2 · · · a n
... ⎩εn/2 1 + δ(ε)εn−2 1/ 2 1/ 2 + o (lnε) 1/ 2 + δ(ε)| lnε| + o εn/2 1 + δ(ε)εn−2 1/ 2 1/ 2 , , if n = 2, if n > 2, (3.42) as ε→0 Finally, due to (3.36), (3.37), and (3.38) we get that 1 ≤ ϕ(ε), − 11 ... Russian Mathematical Surveys, vol 44, no 3, pp 19 5 19 6, 19 89, translation in Uspekhi Matematicheskikh Nauk, vol 44, no 3(267), pp 15 7 15 8, 19 89 [11 ] R R Gadyl’shin and G A Chechkin, “A boundary ... Princeton, NJ, USA, 19 51 [15 ] R R Gadyl’shin, “On the asymptotics of eigenvalues for a periodically fixed membrane,” St Petersburg Mathematical Journal, vol 10 , no 1, pp 1 14 , 19 99, translation...