Question

Inequality and Calculus combo challenge

Answer 1

Let, \(\large{a_1, a_2, a_3, ..., a_n}\) be m (\(\ge\) 1) real numbers which are such that :

\[\large{\sum_{n=1}^{m}a_n \ne 0}\]

Then, prove the inequality:

\[\large{(\sum_{n=1}^{m}n {a_{n}}^2 })/(\sum_{n=1}^{m} a_n)^2 > \cfrac{1}{2\sqrt{m}}\]

Answer 2

Lets invite some users:

@ganeshie8 @Miracrown

Answer 3

@UnkleRhaukus

Answer 4

Give it a try ^^^

It is a very short and moderate difficulty problem

Answer 5

Okay a hint may work

Answer 6

Apply the Cauchy Schwarz inequality to:

\[\large{\sum_{n=1}^{m} {a_n \sqrt{n}\cfrac{1}{\sqrt{n}}}}\]

Answer 7

@vishweshshrimali5 are your questions always that hard.

Answer 8

Uhm.. yes!

Answer 9

But this one is not that difficult.

Answer 10

Okay I am going to share my side of answer

Answer 11

I'll try but don't expect an answer for today. maybe tomorrow.

Answer 12

By the \(\large{\text{Cauchy-Schwarz Inequality}}\) we have:

\[\large{(\sum_{n=1}^{m} a_n)^2 - (\sum_{n=1}^{m} a_n \sqrt{n} \cfrac{1}{\sqrt{n}})^2 \le \sum_{n=1}^{m} n{a_n}^2 \sum_{n=1}^{m} \cfrac{1}{n}} \tag{1}\]

Answer 13

Next, we have, the calculus part:

\[\large{\sum_{n=1}^{m} \cfrac{1}{n} \le 1+ \int_{1}^{m} \cfrac{dx}{x} \le 1+\int_{1}^{m} \cfrac{dx}{\sqrt{x}}}\]

\[\large{= 1+(2\sqrt{m} - 2)) = 2\sqrt{m} - 1 < 2\sqrt{m}}\]

Answer 14

We can easily obtain the required equation using the above inequality and the inequality (1)

Answer 15

\[\large{\color{red}{\text{Hence Proved! - Vishwesh}}}\]

Answer 16

Conclusion - We can always use different branches of maths to arrive at a particular answer.

Answer 17

Interesting, thanks for sharing vishwesh :)

Answer 18

Your welcome @iambatman :)

Answer 19

clever mix of inequality+series+integrals xD

Answer 20

:D I just had the question and the hint and couldn't think of anything else. So, calculus came to my rescue ;)