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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.

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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)

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