Enhanced inequalities about arithmetic and geometric means
EEnhanced inequalities about arithmetic and geometric means
Fang Dai , Li-Gang Xia1. Futian middle school, Ji’an City, Jiangxi Province, China, 343000 Abstract
For n positive numbers ( a k , 1 ≤ k ≤ n ), enhanced inequalities about the arithmetic mean( A n ≡ (cid:80) k a k n ) and the geometric mean ( G n ≡ n √ Π k a k ) are found if some numbers are known,namely, G n A n ≤ ( n − (cid:80) mk =1 r k n − m ) − mn (Π mk =1 r k ) n ≤ , if we know a k = A n r k (1 ≤ k ≤ m ≤ n ) for instance, and G n A n ≤ − mn )Π mk =1 r − n − m k + n (cid:80) mk =1 r k ≤ , if we know a k = G n r k (1 ≤ k ≤ m ≤ n ) for instance. These bounds are better than those derivedfrom S. H. Tung’s work [1]. a r X i v : . [ m a t h . G M ] J u l . INTRODUCTION Let a , a , . . . , a n denote n positive numbers. Let A n be their arithmetic mean, (cid:80) k a k n ,and G n be their geometric mean, n √ Π k a k . We shall prove the following inequalities. G n A n ≤ ( n − (cid:80) mk =1 r k n − m ) − mn (Π mk =1 r k ) n ≤ . (1)if we know a k = A n r k (1 ≤ k ≤ m ≤ n ) for instance. G n A n ≤ − mn )Π mk =1 r − n − m k + n (cid:80) mk =1 r k ≤ a k = G n r k (1 ≤ k ≤ m ≤ n ) for instance.S. H. Tung [1] obtained the following lower bound for A n − G n in terms of the smallestvalue, a , and the largest value, A . A n − G n ≥ ( √ A − √ a ) n . (3)We will see that our results are better than this bound. II. PROOF OF THE INEQUALITIES
Suppose we know the first m numbers, a , a , . . . , a m . We can construct the followinginequality. m (cid:88) k =1 λ n − mk a k + 1Π mi =1 λ i n (cid:88) k = m +1 a k (4)= m (cid:88) k =1 ( λ n − mk − mi =1 λ i ) a k + 1Π mi =1 λ i n (cid:88) k =1 a k (5) ≥ n n (cid:112) Π k a k (6)Suppose we know a k = A n r k ( k = 1 , , . . . , m ). Inserting them into Eq. 5, we obtain anupper bound of G n /A n as a function of λ , λ , . . . , λ m . G n A n ≤ n m (cid:88) k =1 λ n − mk r k + 1Π mk =1 λ k n − (cid:80) mk =1 r k n ≡ f ( λ , λ , . . . , λ m ) . (7) ∂f∂λ i = 0 ( i = 1 , , . . . , m ) gives the best choice, namely, λ i = (cid:18) n − (cid:80) mk =1 r k n − m Π mk =1 r n − m k (cid:19) n r − n − m i , ( i = 1 , , . . . , m ) (8)2nd hence the bound in Ineq. 1.Similarly suppose we know a k = G n r k ( k = 1 , , . . . , m ). Inserting them into Eq. 5, weobtain an upper bound of G n /A n as a function of λ , λ , . . . , λ m . G n A n ≤ mk =1 λ k (1 − n (cid:80) mk =1 r k λ n − mk ) + n (cid:80) mk =1 r k ≡ g ( λ , λ , . . . , λ m ) . (9) ∂f∂λ i = 0 ( i = 1 , , . . . , m ) gives the best choice, namely, λ i = r − n − m i , ( i = 1 , , . . . , m ) (10)and hence the bound in Ineq. 2.For comparison, Tung’s inequality can be written into the following form, G n A n ≤ − n ( √ r − √ r ) , (11)if we know A = A n r and a = A n r with 0 < r ≤ ≤ r ≤ n − r , or G n A n ≤
11 + n ( √ r − √ r ) , (12)if we know A = G n r and a = G n r with 0 < r ≤ ≤ r . We can show that our results arebetter than these bounds using simple calculus. For illustration, different upper bounds of G n /A n as a function of r with r = 5 and n = 10 are compared in Fig. 1. [1] S. H. Tung, Mathematics of computation, Vol. 29, No. 131, pages 834-836, 1975 r00.10.20.30.40.50.60.70.80.91 n / A n U ppe r bound o f G TungThis work r00.10.20.30.40.50.60.70.80.91 n / A n U ppe r bound o f G TungThis work
FIG. 1. Comparison of different upper bounds of G n /A n as a function of r for r = 5 and n = 5. The left diagram correponds to the case of A = A n r and a = A n r while the rightdiagram corresponds to the case of A = G n r and a = G n r . The results derived from Ref. [1] arerepresented by the red-dashed curves (namely, Ineq. 11 and 12. The black curves represent ourresults (namely, Ineq. 1 and 2).. The results derived from Ref. [1] arerepresented by the red-dashed curves (namely, Ineq. 11 and 12. The black curves represent ourresults (namely, Ineq. 1 and 2).