**Given:**

The arithmetic mean is
a1=a+b2
and the geometric mean is
b1=ab
.

In general,
an+1=an+bn2 and bn+1=anbn
.

**Proof: Induction Method**

*Base case:*
n=1

To show that the claim is true when
n=1
.

Consider
a1−b1
, and substitute
a+b2
for
a1
and
ab
for
b1
,

a1−b1=a+b2−ab=12(a+b−2ab)=12((a)2−2⋅a⋅b+(b)2)=12(a−b)2

Since
a>b
, then
a>b
,

(a−b)>0(a−b)2>012(a−b)2>0a1−b1>0 [∵a1−b1=12(a−b)2]

Thus, the inequality is
a1>b1
(1)

Consider
a−a1
and substitute
a+b2
for
a1
.

a−a1=a−a+b2=2a−(a+b)2=2a−a−b2=12(a−b)

Since
a>b
, then
a−b>0
,

12(a−b)>0a−a1>0 [∵a−a1=12(a−b)]

Thus, the inequality is
a>a1
(2)

Consider
b−b1
and substitute
ab
for
b1
,

b−b1=b−ab=b⋅b−a⋅b=b⋅(b−a)

Since
a>b
then
a>b
,

(b−a)<0b(b−a)<0b−b1<0 [∵b−b1=b⋅(b−a)]b<b1

Thus, the inequality is
b1>b
(3)

Combine the equations (1), (2) and (3),
a>a1>b1>b
.

Therefore, the claim is true for
n=1
.

*Induction hypothesis:*
n=k

Assume that the claim is true when
n=k
.

That is, assume
ak>ak+1>bk+1>bk
is true.

*Inductive step:*
n=k+1

To show that the claim is true when
n=k+1
.

That is, to prove that
ak+1>ak+2>bk+2>bk+1
.

Consider
ak+2−bk+2
and substitute
ak+1+bk+12
for
ak+2
and
ak+1bk+1
for
bk+2
,

ak+2−bk+2=ak+1+bk+12−ak+1bk+1=12(ak+1+bk+1−2ak+1bk+1)=12((ak+1)2−2⋅ak+1⋅bk+1+(bk+1)2)=12(ak+1−bk+1)2

From induction hypothesis,
ak+1>bk+1
.

Take square root on both sides,
ak+1>bk+1
.

Therefore, the inequality becomes,

(ak+1−bk+1)>0(ak+1−bk+1)2>012(ak+1−bk+1)2>0ak+2−bk+2>0 [∵ak+2−bk+2=12(ak+1−bk+1)2]

Thus, the inequality is
ak+2>bk+2
(4)

Consider
ak+1−ak+2
and substitute
ak+1+bk+12
for
ak+2
,

ak+1−ak+2=ak+1−ak+1+bk+12=2ak+1−(ak+1+bk+1)2=12(2ak+1−ak+1−bk+1)=12(ak+1−bk+1)

From induction hypothesis
ak+1>bk+1
,

ak+1−ak+2>012(ak+1−ak+2)>0ak+1−ak+2>0 [∵ak+1−ak+2=12(ak+1−ak+2)]

Thus, the inequality is
ak+1>ak+2
(5)

Consider
bk+1−bk+2
and substitute
ak+1bk+1
for
bk+2
,

bk+1−bk+2=bk+1−ak+1bk+1=bk+1⋅bk+1−ak+1⋅bk+1=bk+1(bk+1−ak+1)

From induction hypothesis,
ak+1>bk+1
and
bk+1
is positive.

Take square root on both sides,
ak+1>bk+1
.

Therefore, the inequality becomes,

(bk+1−ak+1)<0bk+1(bk+1−ak+1)<0bk+1−bk+2<0 [∵bk+1−bk+2=bk+1(bk+1−ak+1)]bk+1<bk+2

Thus, the inequality is
bk+2>bk+1
(6)

Combine the equations (4), (5) and (6),
ak+1>ak+2>bk+2>bk+1
.

Hence, by mathematical induction, the claim is true for all *n.*