Inequality -- The Classical Inequality -- A.M. >= G.M.

The Classical Inequality

Arithmetic Mean > Geometrical Mean (A.M.>G.M.)

Let	a₁, a₂, ......, a_n be n positive real numbers
	A.M.= G.M.=
then	A.M. > G.M.
	and the equality holds if and only if a₁= a₂=......= a_n

Lemma 1:	Suppose x > 0 , 0 << 1 then
	See the Proof
Lemma 2:	Let a , b be two positive real numbers
	and ,satisfying < 1 , < 1 and +=1
	then
	See the Proof
Lemma 3:	Suppose that a > 0 , b > 0 , p > 1 , q > 1 and =1
	then
	See the Proof

Proof of Arithmetic Mean > Geometrical Mean

Prove by induction
Consider		=
			by Lemma 2
			by Lemma 2

	where and
Hence by the same technique as before, we can prove by induction,
Assume it is true for n = k
Consider	q₁, q₂, ...... , q_k, q_k₊₁_anda₁, a₂, ...... , a_k, a_k₊₁

	=
	=
By Induction Assumption it is true for n = k,


It is also true for n = 2,

	= a₁q₁ + a₂q₂ +......+ a_kq_k + a_k₊₁q_k₊₁
	a₁q₁ + a₂q₂ +......+ a_nq_n *()**
where	a₁, a₂,......, a_n, q₁, q₂,......, q_n>0
	and	q₁+ q₂+......+ q_n= 1
Let	(i = 1, 2,......, n)
then	q₁+ q₂+......+ q_n= 1
By (*) we get
which is the general case of A.M. > G.M.
Now let	p₁= p₂=......= p_n= 1
we get