Let a and b be the two numbers such that a + b = 6 √ab
⇒ a + b = 3 (2√ab) ⇒

a+b

2√ab

=

3

1

Applying componendo and dividendo, we have,

a+b+2√ab

a+b−2√ab

=

3+1

3−1

⇒

(√a+√b)2

(√a−√b)2

=

4

2

⇒

√a+√b

√a−√b

=

√2

1

Again applying componendo & dividendo, we have

(√a+√b)+(√a−√b)

(√a+√b)−(√a−√b)

=

√2+1

√2−1

⇒

2√a

2√b

=

√2+1

√2−1

⇒

√a

√b

=

√2+1

√2−1

Squaring both sides, we get

a

b

=

(√2+1)2

(√2−1)2

=

2+1+2√2

2+1−2√2

⇒

a

b

=

3+2√2

3−2√2

Hence, proved.