Two inequalities for real numbers

OCTOGON MATHEMATICAL MAGAZINE Vol. 17, No.2, October 2009, pp 742-745 ISSN 1222-5657, ISBN 978-973-88255-5-0, www.hetfalu.ro/octogon

742

Two inequalities for real numbers

Nikolaos Karagiannis ²⁹

We will formulate and prove the following two inequalities:

Theorem

I. If x1, . . . , xn+1>0 such that

n+1Y

i=1

(1 +xi) =

n+1X

i=1

n+1Y

j=1,j6=i

(1 +xj)

then

(n+ 1)ⁿ⁺¹x1· · ·xn+1≤nⁿ⁺¹(1 +x1)· · ·(1 +xn+1) for everyn∈N II. Prove that the following inequality holds

X∞ n=1

n+1X

κ=1

κ²

(2n+ 4)! < 1 20 Proof. I.

n+1Y

i=1

(1 +xi) =

n+1X

i=1 n+1Y

j=1 j6=i

(1 +xj)⇐⇒(1 +x1)· · ·(1 +xn+1) =

= (1 +x2)· · ·(1 +xn+1) +. . .+ (1 +x1)· · ·(1 +xn)⇐⇒

⇐⇒1 = 1

1 +x1 +. . .+ 1

1 +xn+1 ⇐⇒1 = 1 +x1−x1

1 +x1 +. . .

29Received: 16.02.2009

2000Mathematics Subject Classification. 26D15.

Key words and phrases. AM-GM-HM inequality.

Two inequalities for real numbers 743

+1 +x_n+1−x_n+1

1 +x_n+1 ⇐⇒1 = 1− x₁

1 +x₁ +. . .+ 1− x_n+1

1 +x_n+1 ⇐⇒1 =

= (n+ 1)− µ x₁

1 +x₁ +. . .+ x_n+1 1 +x_n+1

¶

⇐⇒n= x₁

1 +x₁ +. . .+ x_n+1 1 +x_n+1 However for the positive real numbers:

1 +x₁

x₁ ,1 +x₂

x₂ , . . . ,1 +x_n+1 x_n+1 from Cauchy’s inequality, one has

n+1

s

(1 +x₁)· · ·(1 +x_n+1)

x₁· · ·x_n+1 ≥ n+ 1

x1

1+x1 +. . .+_1+x^xn+1

n+1

⇐⇒

⇐⇒ ⁿ⁺¹ s

(1 +x₁)· · ·(1 +x_n+1)

x₁· · ·x_n+1 ≥ n+ 1

n ⇐⇒ (1 +x₁)· · ·(1 +x_n+1) x₁· · ·x_n+1 ≥

≥ (n+ 1)ⁿ⁺¹ nⁿ⁺¹ Therefore

(n+ 1)ⁿ⁺¹x₁· · ·x_n+1 ≤nⁿ⁺¹(1 +x₁)· · ·(1 +x_n+1) for everyn∈N Proof.

II. We know that

e^x= 1 + x 1!+x²

2! +. . .+xⁿ

n! +. . .for everyx∈R (1) Thus

e^−x= X∞ n=0

(−x)ⁿ

n! for everyx∈R (2)

From (1) and (2) we obtain:

e^x+e^−x

2 = 1

2 X∞ n=0

µxⁿ

n! + (−1)ⁿxⁿ n!

¶

= 1 +x² 2! +x⁴

4! +. . .= X∞ n=0

x²ⁿ (2n)!

744 Nikolaos Karagiannis Therefore

e^x+e^−x

2 =

X∞ n=0

x²ⁿ

(2n)! for everyx∈R (3)

Differentiating both parts of (3) with respect to x we get e^x−e^−x

2 =

X∞ n=1

2nx²ⁿ⁻¹ (2n)! ⇐⇒

⇐⇒ e^x−e^−x

2 =

X∞ n=0

2(n+ 1)x²ⁿ⁺¹

(2n+ 2)! for everyx∈R (4) Differentiating both parts of (4) with respect to x we get:

e^x+e^−x

2 =

X∞ n=1

2(n+ 1)(2n+ 1)x²ⁿ

(2n+ 2)! for everyx∈R (5) Differentiating both parts of (5) with respect to x we get:

e^x−e^−x

2 =

X∞ n=2

4n(n+ 1)(2n+ 1)

(2n+ 2)! x²ⁿ⁻¹= 24 X∞ n=2

1²+ 2²+. . .+n² (2n+ 2)! x²ⁿ⁻¹ for every x∈R.Thus

e^x−e^−x

48 =

X∞ n=1

1²+ 2²+. . .+ (n+ 1)²

(2n+ 4)! x²ⁿ⁺¹ for every x∈R (6) If we setx= 1, (6) implies :

e−e⁻¹

48 =

X∞ n=1

1²+ 2²+. . .+ (n+ 1)²

(2n+ 4)! = 1²+ 2²

(2·1 + 4)! +1²+ 2²+ 3² (2·2 + 4)! +. . .

= 1

(2·1 + 4)!

X1+1 κ=1

κ²+ 1

(2·2 + 4)!

X2+1 κ=1

κ²+. . .= X∞ n=1

1 (2n+ 4)!

n+1X

κ=1

κ² =

= X∞ n=1

n+1X

κ=1

κ² (2n+ 4)!

Therefore

Two inequalities for real numbers 745

X∞ n=1

n+1X

κ=1

κ²

(2n+ 4)! = e−e⁻¹

48 <0.0489< 1 20. Hence

X∞ n=1

n+1X

κ=1

κ²

(2n+ 4)! < 1 20. REFERENCES

[1] G. H. Hardy, J.E. Littlewood, G. P´olya,Inequalities,Cambridge University Press, 1967.

[2] D.S. Mitrinovi´c, J.E. Pe˘cari´c and A.M. Fink, Inequalities Involving Functions and their Integrals and Derivatives, Kluwer Academic Publishers, Dordrecht, 1991.

[3] D.S. Mitrinovi´c, J.E. Pe˘cari´c and A.M. Fink, Classical New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht, 1993.

[4] Th. M. Rassias, Survey on Classical Inequalities, Kluwer Academic Publishers, Dordrecht, Boston, London, 2000.

Department of Mathematics,

National Technical University of Athens, Zografou Campus, 15780 Athens, Greece E-mail: nikos karagiannis@hotmail.com