aa r X i v : . [ m a t h . G M ] M a r A new proof of the AM-GM-HM inequality
Konstantinos GaitanasMarch 6, 2020
Abstract
In the current note, we present a new, short proof of the famous AM-GM-HM inequalityusing only induction and basic calculus.
Perhaps the most celebrated inequality is the AM-GM-HM inequality which states that if we letAM = a + ... + a n n , GM = n √ a · · · a n , HM = n a + ... + an , then AM ≥ GM ≥ HM holds if a , . . . , a n are positive real numbers. Several authors have provided some novel proofs of this inequality (seefor example [1], [2] or [3] ). The aim of this note is to present a simple new proof which does notseem to appear in the literature. Theorem.
For every a , . . . , a n > , we have a + . . . + a n n ≥ n √ a · · · a n ≥ n a + . . . + a n (1) Proof.
The theorem is obviously true for n = 1 . Suppose now that the theorem holds truefor n − , that is a + ... + a n − n − ≥ n − √ a · · · a n − ≥ n − a + ... + an − . We may write a = a, a = b · a, . . . , a n = b n − · a for some appropriate choice of b , . . . , b n − > . Cancelling a from allsides we can rewrite (1) in the form b + ... + b n − n ≥ n p b · · · b n − ≥ n b + ... + bn − . We first provethe AM-GM inequality: From the induction hypothesis, (multiplying by n − and adding atboth sides) b + . . . + b n − ≥ n − n − p b · · · b n − holds true, so it suffices to show that n − n − p b · · · b n − ≥ n n p b · · · b n − . We let x = n ( n − p b · · · b n − , so the last inequalitycan be written in the form n − x n ≥ nx n − . The function f ( x ) = 1 + ( n − x n − nx n − has a global minimum at x = 1 , the value f (1) = 0 , so f ( x ) ≥ and the inequality holds true for n . In order to prove the GM-HM inequality, we use the induction hypothesis (raising both sides1o the n − n ) so that n p b · · · b n − ≥ ( n − b + ... + bn − ) n − n holds, and it will be enough to prove that ( n − b + . . . + b n − ) n − n ≥ n b + . . . + b n − (2)For this purpose, we let x = b + . . . + b n − and rewrite (2) as ( n − x ) n − n ≥ n x which takinglogarithms of both sides is equivalent to ( n −
1) ln( n −
1) + n ln(1 + x ) − ( n −
1) ln x − n ln n ≥ .Letting g ( x ) = ( n −
1) ln( n −
1) + n ln(1 + x ) − ( n −
1) ln x − n ln n , it is a routine matter toshow that g ( x ) has a global minimum at x = n − , the value g ( n −