Question

Let $x_k$ be real numbers such that $x_k\ge k^4+k^2+1$ for $1\le k\le 2018$. Denote $N=\sum _{k=1}^{2018}k$. Consider the following inequalities :-
I. ${\left(\sum _{k=1}^{2018}k{x}_k\right)}^2\le N\left(\sum _{k=1}^{2018}k{x_k}^2\right)$
II.${\left(\sum _{k=1}^{2018}k{x}_k\right)}^2\le N\left(\sum _{k=1}^{2018}{k}^2{x_k}^2\right)$

• A. both I and II are true
• B. I is true and II is false
• C. I is false and II is true
• D. both I and II are false

Answer 1

The correct option is A both I and II are true

If $x_1,x_2,x_3,.................x_n$ are $n$ numbers then,
${\left(\frac{x_1+x_2+x_3+...........+x_n}{n}\right)}^2\le \left(\frac{{x_1}^2+{x_2}^2+{x_3}^2+............+{x_n}^2}{n}\right)$

Case 1:
Consider
$x_1,x_2,x_2,x_3,x_3,x_3,........................\underset{\text{2018 times}}{\underset{}{x_{2018},x_{2018},......x_{2018}}}$
$\therefore {\left(\frac{x_1+x_2+x_2+...........+x_{2018}+x_{2018}+.........+x_{2018}}{1+2+3+..........+2018}\right)}^2\le \left(\frac{x_1^2+x_2^2+x_2^2+.............+x_{2018}^2+x_{2018}^2+.........x_{2018}^2}{\sum _{k=1}^{2018}k}\right)$
$\Rightarrow {\left(\frac{x_1+2x_2+...........+2018x_{2018}}{\sum _{k=1}^{2018}k}\right)}^2\le \left(\frac{x_1^2+2x_2^2+.............+2018x_{2018}^2}{\sum _{k=1}^{2018}k}\right)$
$\Rightarrow {\left(\sum _{k=1}^{2018}k{x}_k\right)}^2\le \left(\sum _{k=1}^{2018}k\right)\left(\sum _{k=1}^{2018}k{x}_k^2\right)$
$\Rightarrow {\left(\sum _{k=1}^{2018}k{x}_k\right)}^2\le N\left(\sum _{k=1}^{2018}k{x}_k^2\right)$
$\therefore$ Statement I is true.

Case 2:
Consider the numbers,
$x_1,2x_2,3x_3,......,2018x_{2018}$
$\therefore {\left(\frac{x_1+2x_2+...........+2018x_{2018}}{2018}\right)}^2\le \left(\frac{x_1^2+(2x_2{)}^2+.............+(2018x_{2018}{)}^2}{2018}\right)$
$\Rightarrow {(x_1+2x_2+...........2018x_{2018})}^2\le 2018(x_1^2+(2x_2{)}^2+.............+(2018x_{2018}{)}^2)$
$\Rightarrow {\left(\sum _{k=1}^{2018}k{x}_k\right)}^2\le 2018\left(\sum _{k=1}^{2018}{k}^2{x}_k^2\right)$
As we know that,
$N=\sum _{k=1}^{2018}k=\frac{2018\times 2019}{2}>2018$
Hence,
${\left(\sum _{k=1}^{2018}k{x}_k\right)}^2\le N\left(\sum _{k=1}^{2018}{k}^2{x_k}^2\right)$
Statement II is true.