How to prove $(a_1+a_2+\dots a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}\right)\ge n^2$?

Question

let $a_1,a_2,\dots ,a_n$ be a positive real numbers . prove that

$$(a_1+a_2+\dots a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}\right)\ge n^2$$

how to prove this inequality ..is this result prove by Induction

Answer 1

It's just C-S! $$(a_1+a_2+\dots a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}\right)\ge $$ $$\geq\left(\sum_{k=1}^n\left(\sqrt{a_k}\cdot\frac{1}{\sqrt{a_k}}\right)\right)^2=\left(\sum_{k=1}^n1\right)^2=n^2.$$ Also, we can use AM-GM here: $$(a_1+a_2+\dots a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}\right)\geq n\sqrt[n]{\prod_{k=1}^na_k}\cdot\frac{n}{\sqrt[n]{\prod\limits_{k=1}^na_k}}=n^2 .$$ Also, we can use the Tangent Line method.

Indeed, since our inequality is homogeneous, we can assume that $$a_1+a_2+...+a_n=n$$ and we need to prove that $$\sum_{k=1}^n\frac{1}{a_k}\geq n$$ or $$\sum_{k=1}^n\left(\frac{1}{a_k}-1\right)\geq0$$ or $$\sum_{k=1}^n\frac{1-a_k}{a_k}\geq0$$ or $$\sum_{k=1}^n\left(\frac{1-a_k}{a_k}+a_k-1\right)\geq0$$ or $$\sum_{k=1}^n\frac{(a_k-1)^2}{a_k}\geq0.$$ Done!

Answer 2

this is true since $$\frac{a_1+a_2+a_3+...+a_n}{n}\geq \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_n}}$$

Answer 3

Hint: By Cauchy Schwartz inequality we have

$$n=\sum_{i=1}^{n}1= \sum_{i=1}^{n}\frac{1}{\sqrt{a_i}}\sqrt{a_i}\le \left(\sum_{i=1}^{n}\frac{1}{a_i}\right)^{1/2}\left(\sum_{i=1}^{n}a_i\right)^{1/2}$$

Answer 4

This is just the AM-HM inequality. it is easily proved by observing that the following quantity is positive \begin{eqnarray*} \sum_{1 \leq < i < j \leq n} \left( \sqrt{ \frac{a_i}{a_j}}- \sqrt{\frac{a_j}{a_i}} \right) ^2 \geq 0 \end{eqnarray*} Expand this out & add $ \sum_{i=1}^{n} \frac{a_i}{a_i} =n $ to both sides and factorise.

Answer 5

Here is a simple proof: Let $b_i:=1/a_i$ so that $a_ib_i=1$ for all $i$.

Note that $x+1/x\ge1+1$ for $x\gt0$ so apply it to $x=a_ib_j$ for all $i\lt j$ in the following: $(a_1+a_2+\dots+a_n)(b_1+b_2+\dots+b_n)=$ $(1+a_1b_2+\dots+a_1b_n)+(a_2b_1+1+\dots+a_2b_n)+\dots+(a_nb_1+a_nb_2+\dots+1)=$ $(1+1+\dots+1)+(a_1b_2+a_2b_1)+(a_1b_3+a_3b_1)+\dots+(a_1b_n+a_nb_1)+(a_2b_3+a_3b_2)+$ $\dots+(a_2b_n+a_nb_2)+\dots+(a_{n-1}b_n+a_nb_{n-1})\ge n\cdot(1)+n(n-1)/2\cdot(1+1)=n^2.$

You can easily rephrase the above argument as an induction proof if you want to do so.

Answer 6

Another proof: See Art of Problem Solving about Chebyshev's inequality.

There are two different versions of it. Use the second version.

If $a_1\ge a_2\ge \ldots \ge a_n$ and $b_n\ge b_{n-1}\ge \ldots \ge b_1$, then $$n\left(\sum_{i=1}^{n} a_ib_i\right)\le \left(\sum_{i=1}^{n}a_i\right)\left(\sum_{i=1}^{n}b_i\right)$$

In this case, you're given $a_1,a_2,\ldots,a_n>0$. Use the inequality.

Notice that if WLOG (without loss of generality) $a_1\ge a_2\ge \ldots\ge a_n$, then $\frac{1}{a_n}\ge \frac{1}{a_{n-1}}\ge \ldots \ge \frac{1}{a_1}$.