How can I prove that $(a_1+a_2+\dotsb+a_n)(\frac{1}{a_1}+\frac{1}{a_2}+\dotsb+\frac{1}{a_n})\geq n^2$

Question

I've been struggling for several hours, trying to prove this horrible inequality: $(a_1+a_2+\dotsb+a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\dotsb+\frac{1}{a_n}\right)\geq n^2$.

Where each $a_i$'s are positive and $n$ is a natural number.

First I tried the usual "mathematical induction" method, but it made no avail, since I could not show it would be true if $n=k+1$.

Suppose the inequality holds true when $n=k$, i.e.,

$(a_1+a_2+\dotsb+a_k)\left(\frac{1}{a_1}+\frac{1}{a_2}+\dotsb+\frac{1}{a_k}\right)\geq n^2$.

This is true if and only if

$(a_1+a_2+\dotsb+a_k+a_{k+1})\left(\frac{1}{a_1}+\frac{1}{a_2}+\dotsb+\frac{1}{a_k}+\frac{1}{a_{k+1}}\right) -a_{k+1}\left(\frac{1}{a_1}+\dotsb+\frac{1}{a_k}\right)-\frac{1}{a_{k+1}}(a_1+\dotsb+a_k)-\frac{a_{k+1}}{a_{k+1}} \geq n^2$.

And I got stuck here.

The question looks like I have to use AM-GM inequality at some point, but I do not have a clue. Any small hints and clues will be appreciated.

Answer 1

Hint: AM-GM implies $$ a_1+a_2+\cdots +a_n\ge n\sqrt[n]{a_1a_2\cdots a_n} $$ and $$ \frac1{a_1}+\frac1{a_2}+\cdots +\frac1{a_n}\ge \frac{n}{\sqrt[n]{a_1a_2\cdots a_n}}. $$

Answer 2

It is AM-HM inequality $$\frac{a_1+a_2+a_3+...+a_n}{n}\geq \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_n}}$$

Answer 3

Here is the proof by induction that you wanted.

I added a more exact version of the identity used in the proof at the end.

Let $s_n =u_nv_n $ where $u_n=\sum_{k=1}^n a_k, v_n= \sum_{k=1}^n \dfrac1{a_k} $. Then, assuming $s_n \ge n^2$,

$\begin{array}\\ s_{n+1} &=u_{n+1}v_{n+1}\\ &=(u_n+a_{n+1}) (v_n+\dfrac1{a_{n+1}})\\ &=u_nv_n+u_n\dfrac1{a_{n+1}}+a_{n+1}v_n+1\\ &=s_n+u_n\dfrac1{a_{n+1}}+a_{n+1}v_n+1\\ &\ge n^2+u_n\dfrac1{a_{n+1}}+a_{n+1}v_n+1\\ \end{array} $

So it is sufficient to show that $u_n\dfrac1{a_{n+1}}+v_na_{n+1} \ge 2n $.

By simple algebra, if $a, b \ge 0$ then $a+b \ge 2\sqrt{ab} $. (Rewrite as $(\sqrt{a}-\sqrt{b})^2\ge 0$ or, as an identity, $a+b =2\sqrt{ab}+(\sqrt{a}-\sqrt{b})^2$.)

Therefore

$\begin{array}\\ u_n\dfrac1{a_{n+1}}+v_na_{n+1} &\ge \sqrt{(u_n\dfrac1{a_{n+1}})(v_na_{n+1})}\\ &= \sqrt{u_nv_n}\\ &=2\sqrt{s_n}\\ &\ge 2\sqrt{n^2} \qquad\text{by the induction hypothesis}\\ &=2n\\ \end{array} $

and we are done.

I find it interesting that $s_n \ge n^2$ is used twice in the induction step.

Note that, if we use the identity above, $a+b =2\sqrt{ab}+(\sqrt{a}-\sqrt{b})^2$, we get this:

$\begin{array}\\ s_{n+1} &=s_n+u_n\dfrac1{a_{n+1}}+a_{n+1}v_n+1\\ &=s_n+2\sqrt{u_n\dfrac1{a_{n+1}}a_{n+1}v_n}+1+(\sqrt{u_n\dfrac1{a_{n+1}}}-\sqrt{a_{n+1}v_n})^2\\ &=s_n+2\sqrt{s_n}+1+\dfrac1{a_{n+1}}(\sqrt{u_n}-a_{n+1}\sqrt{v_n})^2\\ &=(\sqrt{s_n}+1)^2+\dfrac1{a_{n+1}}(\sqrt{u_n}-a_{n+1}\sqrt{v_n})^2\\ &\ge(\sqrt{s_n}+1)^2\\ \end{array} $

with equality if and only if $a_{n+1} =\sqrt{\dfrac{u_n}{v_n}} =\sqrt{\dfrac{\sum_{k=1}^n a_k}{\sum_{k=1}^n \dfrac1{a_k}}} $.