Let a and b be positive numbers with a > b. Let $a_1$ be their arithmetic mean and $b_1$ their geometric mean:
$$ a_1 = \frac{a + b}{2}$$ $$b_1 = \sqrt{a*b} $$
Repeat this process so that, in general: $$ a_{n+1} = \frac{a_n + b_n}{2}$$ $$b_{n+1} = \sqrt{a_n*b_n} $$
Use mathematical induction to show that: $$ a_n>a_{n+1}>b_{n+1}>b_n$$
My solution: First we need to prove the base case(n = 1).
First i will show, that $a_1 > b_1$: $$ \frac{a+b}{2} > \sqrt{a*b} $$ With algebraic manipulation we become $$ (a - b)^2 > 0$$ and that is true, because a > b.
Second, i will show, that $a_1 > a_2$ $$a_1 > \frac{a_1 + b_1}{2}$$ by multiplying both sides by 2 we become $a_1 - b_1 >0$ which i true(from my first part).
Thirdly, i will show, that $ a_2 > b_2 $ $$\frac{a_1 + b_1}{2} > \sqrt {a_1 *b_1} $$
By raising to square and multiplying by 4 we become: $(a_1 - b_1)^2 >0$ which is true.
And finally we can show that $b_2 > b_1$ just by raising both sides to the 2 power.
My question: how can we prove in induction step, that $a_{n+1} > b_{n+1}$ ? If we will prove this fact, rest parts are easy. Many thanks.
Suppose $a_{n-1}>a_n>b_{n}>b_{n-1}$
Then $a_{n+1}=\frac{a_n+b_n}{2}$
So $a_n>a_{n+1}>b_n$ (the mean is greater than the smaller number, smaller than the larger number)
$b_{n+1}=\sqrt{a_n b_n}$
You can prove $a_{n+1}>b_{n+1}$ (AM>GM) by observing $(a_{n}-b_{n})^2>0$ (since $a_{n}\neq b_{n}$)
$a_{n}^2-2a_{n}b_{n}+b_{n}^2>0$
Add $4a_{n}b_{n}$ to both sides
$(a_{n}+b_{n})^2>4a_{n}b_{n}$
$\frac{a_{n}+b_{n}}{2}>\sqrt{a_{n}b_{n}}$ (We can take the square root because both sides are positive)