How to prove that $a_nb_n < a_n+b_n$

Question

$a_n, b_n \in \mathbb R; n \in \mathbb N; x \in \mathbb R \land x > 1$ are defined as:

$$a_1 = 1 \land b_1 = x$$ $$a_{n+1} = \frac {2a_nb_n} {a_n+b_n} \land b_{n+1} = \frac {a_n+b_n} 2$$

How can I prove that $a_nb_n < a_n+b_n$ ?

Answer 1

Your claim is true if and only if $x\le 4$.

We will use induction to show some facts about both sequences. Namely,

$$a_{n+1}b_{n+1}=\frac{2a_nb_n}{a_n+b_n}\cdot\frac{a_n+b_n}2=a_nb_n=x$$

$$b_{n+1}-a_{n+1}=\frac{a_n+b_n}2-\frac{2a_nb_n}{a_n+b_n}=\frac{(b_n-a_n)^2}{2(a_n+b_n)}>0$$

Also, $a_{n+1}>a_n$: $$a_{n+1}-a_n=\frac{2x}{a_n+b_n}-a_n>\frac{2x}{2b_n}-a_n=a_n-a_n=0$$

Since $a_nb_n$ is constant and $a_n$ is strictly increasing, $b_n$ is strictly decreasing.

All this implies that $a=\lim a_n$ and $b=\lim b_n$ exist. Also, $a_n<a\le b<b_n$ for every $n\in\Bbb N$.

Taking limits in the equations: $$a=\frac{2ab}{a+b}$$ $$b=\frac{a+b}2$$ And then $$2(a+b)^2=4ab+(a+b)^2$$ That is $$(a-b)^2=0$$ Then $a=b$. So $a=b=\sqrt x$.

Avoiding formalism for the moment, for $n$ large enough, we see that $a_n+b_n$ will be near from $2\sqrt x$, which, for $x> 4$, is lesser than $x$.

Assuming $x\le 4$, by GM-AM inequality, $$\frac{a_n+b_n}{a_nb_n}>\frac{2\sqrt x}x=\frac2{\sqrt x}\ge 1$$

But if $x>4$, for any $\epsilon>0$ there exists some $n\in\Bbb N$ such that $$\sqrt x-\epsilon<a_n<\sqrt x<b_n<\sqrt x+\epsilon$$ and then $$a_n+b_n<2\sqrt x+\epsilon$$ Taking $\epsilon=x-2\sqrt x>0$, we see that the claim is false.

Answer 2

$\frac{2ab}{a+b}$ is the harmonic mean of $a$ and $b$ while $\frac{a+b}2$ is the arithmetic mean. When $a,b\gt0$, the harmonic mean is less than or equal to the geometric mean $\left(\sqrt{ab}\right)$, which is less than or equal to the arithmetic mean.

It is useful to note that the harmonic mean of $x$ and $\frac1x$ is $\frac{2x}{x^2+1}$ while the arithmetic mean of $x$ and $\frac1x$ is $\frac{x^2+1}{2x}$. Thus, if $a_n$ and $b_n$ are reciprocals, $a_{n+1}$ and $b_{n+1}$ are also reciprocals. In this case, $a_n$ approaches $1$ from below and $b_n$ approaches $1$ from above. Since the mean (harmonic, geometric, and arithmetic) of a multiple of two numbers is that multiple of the mean of the same two numbers, we get that $$ a_n\nearrow\sqrt{a_1b_1}\tag{1} $$ and $$ b_n\searrow\sqrt{a_1b_1}\tag{2} $$ From $(1)$, we see that $a_n\le2$ whenever $x\le4$. In this case, $$ \frac{a_nb_n}{a_n+b_n}=\frac{a_{n+1}}{2}\le1\tag{3} $$ If $x\gt4$, then at some point, $a_{n+1}\gt2$ and then $$ \frac{a_nb_n}{a_n+b_n}=\frac{a_{n+1}}{2}\gt1\tag{4} $$