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Proof for AM-GM inequality using calculus tehniques

Dorin M˘arghidanu 1

There are many proofs for the well known inequality between arithmetic mean and geometric
mean, or Cauchy’s inequality. See [1] and the more extensive bibliography contained therein.
In this paper we shall give a new one. For this, we will give two preliminaries calculus results.

Although the first result is known, we recall it - as a basis for the next.
Lemma 1. For x > 1, holds the following inequality, e 1− 1 x < x<e x−1 .
Proof. We consider the function f : [1, ∞) → [1, ∞), f(x) = ln x and apply Lagrange’s theorem
lnx 1
on the interval [1, x] ⊂ [1, ∞). There is c ∈ (1, x) so that = .
x−1 c
1 1 1 lnx x−1
Since < <1, it comes < <1, thus <ln x<x−1.
x c x x−1 x
By exponentiation we get indeed the inequality stated earlier.
An inequality - interesting in itself - is stated in.

b−a b b−a
Lemma 2. For 0 < a ≤ b, occurs the following e b ≤ ≤e a .
a
a
Proof. For a = b, we have obviously an equality. Supposing that b > a, then with x= >1 in
b
Lemma 1, it comes immediately the double inequality in Lemma 2.
∗
∗
Theorem 1. (Cauchy’s inequality or AM-GM inequality.) If x 1 , x 2 , . . . , x n ∈ R , n ∈ N and
+
√
x 1 + x 2 + . . . + x n
A := , G := n x 1 · x 2 · . . . · x n , then A ≥ G, with equality if and only if
n
x 1 = x 2 = . . . = x n .
∗
Proof. For h ∈ N , we note: P h := x 1 · x 2 · . . . · x h , S h := x 1 + x 2 + . . . + x h ; thus A = S n ,
√ n
G := n P n . Without restricting the generality, we suppose that 0 < x 1 ≤ x 2 ≤ . . . ≤ x n . Since
∗
min{x 1 , x 2 , . . . , x n } ≤ G ≤ max{x 1 , x 2 , . . . , x n } results that exist k ∈ N so that
0 < x 1 ≤ x 2 ≤ . . . ≤ x n ≤ G ≤ x k+1 ≤ . . . ≤ x n (1)
By it’s very name – of (geometric) mean, G justifies it’s intermediate position!). With (1) and
the left inequality in Lemma 2, we have:
G
G−x 1
x 1 ≤ G → e G ≤ (2 1 )
x 1
G
G−x 2
x 2 ≤ G → e G ≤ (2 2 )
x 2
. . .

G−x k G
x k ≤ G → e G ≤ (2 k )
x k
1
Profesor dr., Colegiul Nat , ional ,,Al. I. Cuza”, Corabia, d.marghidanu@gmail.com

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